Proof (a) is immediate from (ci) and (cii). For (b), see the proof in [Reference Damnjanovic3, 2.1(b)]; (c) and (d) follow from (b).

Proof Assume that ${\varSigma}^{\ast}\models \mathrm{x}={\mathrm{sax}}_1\dots {\mathrm{x}}_{\mathrm{n}}\ \&\ \mathrm{y}={\mathrm{tay}}_1\dots {\mathrm{y}}_{\mathrm{n}}$ , where ${\varSigma}^{\ast}\models {\mathrm{Tally}}_{\mathrm{b}}(\mathrm{s})\ \&\ {\mathrm{Tally}}_{\mathrm{b}}(\mathrm{t})$ and . Suppose ${\varSigma}^{\ast}\models \mathrm{x}=\mathrm{y}$ . Then ${\varSigma}^{\ast}\models {\mathrm{sax}}_1\dots {\mathrm{x}}_{\mathrm{n}}={\mathrm{tay}}_1\dots {\mathrm{y}}_{\mathrm{n}}$ , whence ${\varSigma}^{\ast}\models \mathrm{sBx}\ \& \ \mathrm{tBx}$ . In that case, we have that ${\varSigma}^{\ast}\models \mathrm{sBt}\;\mathrm{v}\;\mathrm{s}=\mathrm{t}\;\mathrm{v}\;\mathrm{t}\mathrm{Bs}$ . Suppose, for a reductio, that ${\varSigma}^{\ast}\models \mathrm{sBt}$ . Then ${\varSigma}^{\ast}\models \exists \mathrm{u}\;({\mathrm{Tally}}_{\mathrm{b}}(\mathrm{u})\ \& \ \mathrm{t}=\mathrm{su})$ , whence ${\varSigma}^{\ast}\models {\mathrm{sax}}_1\dots {\mathrm{x}}_{\mathrm{n}}={\mathrm{suay}}_1\dots {\mathrm{y}}_{\mathrm{n}}$ . But then ${\varSigma}^{\ast}\models {\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{n}}={\mathrm{uay}}_1\dots {\mathrm{y}}_{\mathrm{n}}$ , which is a contradiction because ${\varSigma}^{\ast}\models {\mathrm{Tally}}_{\mathrm{b}}(\mathrm{u})$ . Hence ${\varSigma}^{\ast}\models \neg \mathrm{sBt}$ . Exactly analogously, ${\varSigma}^{\ast}\models \neg \mathrm{tBs}$ . It follows that ${\varSigma}^{\ast}\models \mathrm{s}=\mathrm{t}$ .

Thus we obtain that ${\varSigma}^{\ast}\models {\mathrm{sax}}_1\dots {\mathrm{x}}_{\mathrm{n}}={\mathrm{say}}_1\dots {\mathrm{y}}_{\mathrm{n}}$ , whence ${\varSigma}^{\ast}\models {\mathrm{x}}_1\dots {\mathrm{x}}_{\mathrm{n}}=\mathrm{z}={\mathrm{y}}_1\dots {\mathrm{y}}_{\mathrm{n}}$ . So ${\varSigma}^{\ast}\models {\mathrm{x}}_1\mathrm{Bz}\ \&\ {\mathrm{y}}_1\mathrm{Bz}$ . Then ${\varSigma}^{\ast}\models {\mathrm{x}}_1{\mathrm{By}}_1\mathrm{v}\;{\mathrm{x}}_1={\mathrm{y}}_1\;\mathrm{v}\;{\mathrm{y}}_1{\mathrm{Bx}}_1$ . But from , by 2.3(a), we have ${\varSigma}^{\ast}\models \neg {\mathrm{x}}_1{\mathrm{By}}_1\ \&\ \neg {\mathrm{y}}_1{\mathrm{Bx}}_1$ . Therefore, ${\varSigma}^{\ast}\models {\mathrm{x}}_1={\mathrm{y}}_1$ . Assume now that ${\varSigma}^{\ast }{\models}\wedge_{1\le \mathrm{i}\le \mathrm{k}}{\mathrm{x}}_{\mathrm{i}}={\mathrm{y}}_{\mathrm{i}}$ for $\mathrm{k}<\mathrm{n}$ . Then from ${\varSigma}^{\ast}\models \mathrm{s}=\mathrm{t}$ and hypothesis ${\varSigma}^{\ast}\models {\mathrm{sax}}_1\dots {\mathrm{x}}_{\mathrm{n}}={\mathrm{tay}}_1\dots {\mathrm{y}}_{\mathrm{n}}$ we obtain

$$\begin{align*}{\varSigma}^{\ast}\models {\mathrm{sax}}_1\dots {\mathrm{x}}_{\mathrm{k}}({\mathrm{x}}_{\mathrm{k}+1}\dots {\mathrm{x}}_{\mathrm{n}})={\mathrm{sax}}_1\dots {\mathrm{x}}_{\mathrm{k}}({\mathrm{y}}_{\mathrm{k}+1}\dots {\mathrm{y}}_{\mathrm{n}}),\end{align*}$$

whence ${\varSigma}^{\ast}\models {\mathrm{x}}_{\mathrm{k}+1}\dots {\mathrm{x}}_{\mathrm{n}}={\mathrm{y}}_{\mathrm{k}+1}\dots {\mathrm{y}}_{\mathrm{n}}$ . Then the same argument as above shows that ${\varSigma}^{\ast}\models {\mathrm{x}}_{\mathrm{k}+1}={\mathrm{y}}_{\mathrm{k}+1}$ . Hence for all $\mathrm{j}$ , $1\le \mathrm{j}\le \mathrm{n}$ , ${\varSigma}^{\ast}\models {\mathrm{x}}_{\mathrm{j}}={\mathrm{y}}_{\mathrm{j}}$ .

Proof For (a), assume . By 2.2, ${\varSigma}^{\ast}\models \mathrm{bBx}\ \&\ \mathrm{aaEx}$ . Hence ${\varSigma}^{\ast}\models \exists {\mathrm{x}}_1\mathrm{x}={\mathrm{x}}_1\mathrm{aa}$ . So $\mathrm{x}$ has at least one proper initial segment $(={\mathrm{x}}_1\mathrm{a})$ that ends with $\mathrm{a}$ . Let ${\mathrm{s}}_0$ be the shortest such initial segment. Then ${\varSigma}^{\ast}\models {\mathrm{aEs}}_0$ , so ${\varSigma}^{\ast}\models \exists \mathrm{t}\;{\mathrm{s}}_0=\mathrm{ta}$ . But then ${\varSigma}^{\ast}\models {\mathrm{s}}_0\mathrm{Bx}\ \&\ {\mathrm{bBs}}_0$ , so we must have ${\varSigma}^{\ast}\models {\mathrm{Tally}}_{\mathrm{b}}(\mathrm{t})$ by the choice of ${\mathrm{s}}_0$ .

For (b), assume ${\varSigma}^{\ast}\models \mathrm{x}=\mathrm{bay}$ where . Then

$$\begin{align*}{\varSigma}^{\ast}\models \alpha (\mathrm{y})=\alpha (\mathrm{bay}){-}\underline{1}=(\beta (\mathrm{bay})+\underline{1}){-}\underline{1}=\beta (\mathrm{bay})=\beta (\mathrm{y})+\underline{1}.\end{align*}$$

Suppose ${\varSigma}^{\ast}\models \mathrm{zBy}$ . Then ${\varSigma}^{\ast}\models (\mathrm{baz})\mathrm{B}(\mathrm{bay})$ . Hence from hypothesis we have ${\varSigma}^{\ast}\models \alpha (\mathrm{z})=\alpha (\mathrm{baz})\hbox{--} \underline{1}\le \beta (\mathrm{baz})\hbox{--} \underline{1}=(\beta (\mathrm{z})+\underline{1})\hbox{--} \underline{1} =\beta (\mathrm{z})$ . Hence also ${\varSigma}^{\ast}\models \forall \mathrm{z}\;(\mathrm{z}\mathrm{By}\to \alpha (\mathrm{z})\le \beta (\mathrm{z}))$ . Therefore, .

Proof Assume . Then, by 2.5(a), ${\varSigma}^{\ast}\models \exists \mathrm{y}\;\mathrm{u}=\mathrm{tay}$ where ${\varSigma}^{\ast}\models {\mathrm{Tally}}_{\mathrm{b}}(\mathrm{t})$ . Let $\mathrm{t}={\mathrm{b}}^{\mathrm{n}}$ . If $\mathrm{n}=1$ , we are done, for we can let ${\varSigma}^{\ast}\models {\mathrm{u}}_1=\mathrm{y}$ .

Suppose $\mathrm{n}>1$ . We claim that if ${\varSigma}^{\ast}\models \mathrm{u}={\mathrm{b}}^{\mathrm{n}}\mathrm{ay}$ , then ${\varSigma}^{\ast}\models \alpha (\mathrm{y})+\beta (\mathrm{y})\ge \underline {\mathrm{n}}$ .

We have that

$$\begin{align*}{\varSigma}^{\ast}\models \alpha (\mathrm{u})=\alpha ({\mathrm{b}}^{\mathrm{n}}\mathrm{ay})=\underline{1}+\alpha (\mathrm{y})\kern0.24em \mathrm{and}\;{\varSigma}^{\ast}\models \beta (\mathrm{u})=\beta ({\mathrm{b}}^{\mathrm{n}}\mathrm{ay})=\underline {\mathrm{n}}+\beta (\mathrm{y}).\end{align*}$$

From , we have ${\varSigma}^{\ast}\models \alpha (\mathrm{u})=\beta (\mathrm{u})+\underline{1}$ .

Then ${\varSigma}^{\ast}\models \underline{1}+\alpha (\mathrm{y})=\mathrm{n}+\beta (\mathrm{y})+1$ . So ${\varSigma}^{\ast}\models \alpha (\mathrm{y})=\beta (\mathrm{y})+\underline {\mathrm{n}}\ge \underline {\mathrm{n}}$ . But then ${\varSigma}^{\ast}\models \alpha (\mathrm{y})+\beta (\mathrm{y})\ge \underline {\mathrm{n}}+\beta (\mathrm{y})\ge \underline {\mathrm{n}}$ , which proves the claim.

Now let ${\varSigma}^{\ast}\models \alpha (\mathrm{y})+\beta (\mathrm{y})=\underline {\mathrm{k}}\ge \underline {\mathrm{n}}$ . By 2.1 we have that

$$\begin{align*}{\varSigma}^{\ast}\models \exists {\mathrm{y}}_1\dots {\mathrm{y}}_{\mathrm{k}}(\wedge_{1\le \mathrm{i}\le \mathrm{k}}\mathrm{Digit}({\mathrm{y}}_{\mathrm{i}})\ \&\ \mathrm{y}={\mathrm{y}}_1\dots {\mathrm{y}}_{\mathrm{k}}).\end{align*}$$

If $\mathrm{k}=\mathrm{n}$ , we are done. If $\mathrm{k}>\mathrm{n}$ , then ${\varSigma}^{\ast}\models \exists \mathrm{z}\;\mathrm{y}={\mathrm{y}}_1\dots {\mathrm{y}}_{\mathrm{n}}\mathrm{z}={\mathrm{y}}_1\dots {\mathrm{y}}_{\mathrm{n}-1}({\mathrm{y}}_{\mathrm{n}}\mathrm{z})$ .

But then ${\varSigma}^{\ast}\models \mathrm{u}={\mathrm{b}}^{\mathrm{n}}\mathrm{ay}={\mathrm{b}}^{\mathrm{n}}{\mathrm{au}}_1\dots {\mathrm{u}}_{\mathrm{n}}$ where ${\varSigma}^{\ast }{\models}\wedge_{1\le \mathrm{i}<\mathrm{n}}\;{\mathrm{u}}_{\mathrm{i}}={\mathrm{y}}_{\mathrm{i}}\ \&\ {\mathrm{u}}_{\mathrm{n}}={\mathrm{y}}_{\mathrm{n}}\mathrm{z}$ .

Proof Assume , and let ${\varSigma}^{\ast}\models \mathrm{y}={\mathrm{b}}^{\mathrm{n}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{n}}$ . Then

$$\begin{align*} \begin{array}{l}{\varSigma}^{\ast}\models \alpha (\mathrm{y})=\alpha ({\mathrm{b}}^{\mathrm{n}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{n}})=\underline{1}+{\sum}_{1\le \mathrm{i}\le \mathrm{n}}\alpha ({\mathrm{x}}_{\mathrm{i}})=\underline{1}+{\sum}_{1\le \mathrm{i}\le \mathrm{n}}(\beta ({\mathrm{x}}_{\mathrm{i}})+\underline{1})=\\ {}\kern0.84em =\underline{1}+\underline {\mathrm{n}}+{\sum}_{1\le \mathrm{i}\le \mathrm{n}}\beta ({\mathrm{x}}_{\mathrm{i}})=\beta ({\mathrm{b}}^{\mathrm{n}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{n}})+\underline{1}=\beta (\mathrm{y})+\underline{1}.\end{array}\end{align*}$$

Hence (ci) holds. Suppose now that ${\varSigma}^{\ast}\models \mathrm{wBy}$ . We have that

$$\begin{align*} \begin{array}{l}{\varSigma}^{\ast}\models \mathrm{y}={\mathrm{b}}^{\mathrm{n}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{n}}\ \&\ \mathrm{w}\mathrm{By}\to {\mathrm{wBb}}^{\mathrm{n}}\;\mathrm{v}\;\mathrm{w}={\mathrm{b}}^{\mathrm{n}}\;\mathrm{v}\;\mathrm{w}={\mathrm{b}}^{\mathrm{n}}\mathrm{a}\;\mathrm{v}\\[2pt] \hspace{9pt}\qquad\qquad{}\mathrm{v}\;(\exists {\mathrm{z}}_1({\mathrm{z}}_1{\mathrm{Bx}}_1\ \&\ \mathrm{w}={\mathrm{b}}^{\mathrm{n}}{\mathrm{az}}_1)\;\mathrm{v}\kern0.24em \mathrm{w}={\mathrm{b}}^{\mathrm{n}}{\mathrm{ax}}_1)\;\mathrm{v}\\[2pt] \hspace{-9pt}\mathrm{v}\;{\vee}_{1<\mathrm{i}<\mathrm{n}}(\exists {\mathrm{z}}_{\mathrm{i}}({\mathrm{z}}_{\mathrm{i}}{\mathrm{Bx}}_{\mathrm{i}}\ \&\ \mathrm{w}={\mathrm{b}}^{\mathrm{n}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{i}-1}{\mathrm{z}}_{\mathrm{i}})\;\mathrm{v}\;\mathrm{w}={\mathrm{b}}^{\mathrm{n}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{i}})\;\mathrm{v}\\[2pt] \qquad\qquad\qquad\qquad\quad\qquad{}\mathrm{v}\;\exists {\mathrm{z}}_{\mathrm{n}}({\mathrm{z}}_{\mathrm{n}}{\mathrm{Bx}}_{\mathrm{n}}\ \&\ \mathrm{w}={\mathrm{b}}^{\mathrm{n}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{n}-1}{\mathrm{z}}_{\mathrm{n}}).\end{array}\end{align*}$$

We consider one of the cases as an illustration of the proof. Suppose

$$\begin{align*}{\varSigma}^{\ast}\models \exists {\mathrm{z}}_{\mathrm{i}}({\mathrm{z}}_{\mathrm{i}}{\mathrm{Bx}}_{\mathrm{i}}\ \&\ \mathrm{w}={\mathrm{b}}^{\mathrm{n}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{i}-1}{\mathrm{z}}_{\mathrm{i}})\kern0.36em \mathrm{where}\kern0.36em 1<\mathrm{i}<\mathrm{n}.\end{align*}$$

From we have ${\varSigma}^{\ast }{\models}\wedge_{1\le \mathrm{j}<\mathrm{i}}\alpha ({\mathrm{x}}_{\mathrm{j}})=\beta ({\mathrm{x}}_{\mathrm{j}})+\underline{1}$ . On the other hand, from , we have ${\varSigma}^{\ast}\models \alpha ({\mathrm{z}}_{\mathrm{i}})\le \beta ({\mathrm{z}}_{\mathrm{i}})$ . Then

$$\begin{align*} \begin{array}{l}\hspace{-8pt}{\varSigma}^{\ast}\models \alpha (\mathrm{w})=\alpha ({\mathrm{b}}^{\mathrm{n}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{i}-1}{\mathrm{z}}_{\mathrm{i}})=\underline{1}+{\sum}_{1\le \mathrm{j}<\mathrm{i}}\alpha ({\mathrm{x}}_{\mathrm{j}})+\alpha ({\mathrm{z}}_{\mathrm{i}})=\\ {}\hspace{45pt} =\underline{1}+{\sum}_{1\le \mathrm{j}<\mathrm{i}}(\beta ({\mathrm{x}}_{\mathrm{j}})+\underline{1})+\alpha ({\mathrm{z}}_{\mathrm{i}})\le {\sum}_{1\le \mathrm{j}<\mathrm{i}}\beta ({\mathrm{x}}_{\mathrm{j}})+\underline {\mathrm{i}}+\beta ({\mathrm{z}}_{\mathrm{i}})\le \\ {}\hspace{45pt} \le \underline {\mathrm{n}}+{\sum}_{1\le \mathrm{j}<\mathrm{i}}\beta ({\mathrm{x}}_{\mathrm{j}})+\beta ({\mathrm{z}}_{\mathrm{i}})=\beta ({\mathrm{b}}^{\mathrm{n}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{i}-1}{\mathrm{z}}_{\mathrm{i}})=\beta (\mathrm{w}),\end{array}\end{align*}$$

as required. Detailed consideration of the cases shows that ${\varSigma}^{\ast}\models \forall \mathrm{w}\;(\mathrm{w}\mathrm{By}\to \alpha (\mathrm{w})\le \beta (\mathrm{w}))$ . This establishes (cii), and so .

Proof We have from ${\varSigma}^{\ast}\models {\mathrm{b}}^{\mathrm{n}}{\mathrm{ay}}_1\dots {\mathrm{y}}_{\mathrm{n}}=\mathrm{z}={\mathrm{x}}_1\mathrm{aa}$ that ${\varSigma}^{\ast}\models {\mathrm{b}}^{\mathrm{n}}\mathrm{Bz}\ \& {\mathrm{x}}_1\mathrm{Bz}$ . Then ${\varSigma}^{\ast}\models {\mathrm{b}}^{\mathrm{n}}{\mathrm{Bx}}_1\mathrm{v}\;{\mathrm{b}}^{\mathrm{n}}={\mathrm{x}}_1\;\mathrm{v}\;{\mathrm{x}}_1{\mathrm{Bb}}^{\mathrm{n}}$ . Suppose, for a reductio, that ${\varSigma}^{\ast}\models {\mathrm{x}}_1{\mathrm{Bb}}^{\mathrm{n}}$ . Then ${\varSigma}^{\ast}\models \exists {\mathrm{z}}_1{\mathrm{b}}^{\mathrm{n}}={\mathrm{x}}_1{\mathrm{z}}_1$ where ${\varSigma}^{\ast}\models {\mathrm{Tally}}_{\mathrm{b}}({\mathrm{z}}_1)$ . But then ${\varSigma}^{\ast}\models ({\mathrm{x}}_1{\mathrm{z}}_1){\mathrm{ay}}_1\dots {\mathrm{y}}_{\mathrm{n}}={\mathrm{x}}_1\mathrm{aa}$ , that is, ${\varSigma}^{\ast}\models {\mathrm{x}}_1({\mathrm{z}}_1{\mathrm{ay}}_1\dots {\mathrm{y}}_{\mathrm{n}})={\mathrm{x}}_1\mathrm{aa}$ , whence ${\varSigma}^{\ast}\models {\mathrm{z}}_1{\mathrm{ay}}_1\dots {\mathrm{y}}_{\mathrm{n}}=\mathrm{aa}$ , which is a contradiction because ${\varSigma}^{\ast}\models {\mathrm{Tally}}_{\mathrm{b}}({\mathrm{z}}_1)$ . Therefore, ${\varSigma}^{\ast}\models {\mathrm{b}}^{\mathrm{n}}{\mathrm{Bx}}_1\mathrm{v}\kern0.24em {\mathrm{b}}^{\mathrm{n}}={\mathrm{x}}_1$ .

Proof Assume . By 2.6 we have that ${\varSigma}^{\ast}\models \exists {\mathrm{y}}_1\dots {\mathrm{y}}_{\mathrm{n}}\mathrm{u}={\mathrm{b}}^{\mathrm{n}}{\mathrm{ay}}_1\dots {\mathrm{y}}_{\mathrm{n}}$ . We need to show that there are unique strings ${\mathrm{u}}_1,\dots, {\mathrm{u}}_{\mathrm{n}}$ such that ${\varSigma}^{\ast}\models {\mathrm{u}}_1\dots {\mathrm{u}}_{\mathrm{n}}={\mathrm{y}}_1\dots {\mathrm{y}}_{\mathrm{n}}$ .

If $\mathrm{n}=1$ , then ${\varSigma}^{\ast}\models \mathrm{u}=\mathrm{bay}$ . From hypothesis we have by 2.5(b) that . So we may let ${\mathrm{u}}_1=\mathrm{y}$ . Uniqueness is immediate.

So we may assume that $\mathrm{n}\ge 2$ . By 2.2, from we have that ${\varSigma}^{\ast}\models \mathrm{aaEu}$ . Hence ${\varSigma}^{\ast}\models \exists {\mathrm{x}}_1\mathrm{u}={\mathrm{x}}_1\mathrm{aa}$ . By 2.10, we have that ${\varSigma}^{\ast}\models {\mathrm{x}}_1={\mathrm{b}}^{\mathrm{n}}\;\mathrm{v}\;{\mathrm{b}}^{\mathrm{n}}{\mathrm{Bx}}_1$ . Now, we cannot have ${\varSigma}^{\ast}\models {\mathrm{x}}_1={\mathrm{b}}^{\mathrm{n}}$ , for then ${\varSigma}^{\ast}\models {\mathrm{b}}^{\mathrm{n}}{\mathrm{ay}}_1\dots {\mathrm{y}}_{\mathrm{n}}={\mathrm{b}}^{\mathrm{n}}\mathrm{aa}$ , whence ${\varSigma}^{\ast}\models {\mathrm{y}}_1\dots {\mathrm{y}}_{\mathrm{n}}=\mathrm{a}$ , which is impossible given that $\mathrm{n}\ge 2$ . Therefore, ${\varSigma}^{\ast}\models {\mathrm{b}}^{\mathrm{n}}{\mathrm{Bx}}_1$ . Then ${\varSigma}^{\ast}\models \exists {\mathrm{x}}_2{\mathrm{x}}_1={\mathrm{b}}^{\mathrm{n}}{\mathrm{x}}_2$ , whence

$$\begin{align*}{\varSigma}^{\ast}\models {\mathrm{b}}^{\mathrm{n}}{\mathrm{ay}}_1\dots {\mathrm{y}}_{\mathrm{n}}=\mathrm{u}={\mathrm{x}}_1\mathrm{aa}=({\mathrm{b}}^{\mathrm{n}}{\mathrm{x}}_2)\mathrm{aa}={\mathrm{b}}^{\mathrm{n}}({\mathrm{x}}_2\mathrm{aa}).\end{align*}$$

Then ${\varSigma}^{\ast}\models {\mathrm{ay}}_1\dots {\mathrm{y}}_{\mathrm{n}}={\mathrm{x}}_2\mathrm{aa}$ , and further, ${\varSigma}^{\ast}\models {\mathrm{x}}_2=\mathrm{a}\;\mathrm{v}\;{\mathrm{aBx}}_2$ .

If ${\varSigma}^{\ast}\models {\mathrm{x}}_2=\mathrm{a}$ , then ${\varSigma}^{\ast}\models {\mathrm{ay}}_1\dots {\mathrm{y}}_{\mathrm{n}}=\mathrm{aaa}$ , so ${\varSigma}^{\ast}\models {\mathrm{y}}_1\dots {\mathrm{y}}_{\mathrm{n}}=\mathrm{aa}$ . But then $\mathrm{n}=2$ and ${\varSigma}^{\ast}\models {\mathrm{y}}_1={\mathrm{y}}_2=\mathrm{a}$ . If we let ${\mathrm{u}}_1={\mathrm{u}}_2=\mathrm{a}$ we have that and uniqueness is immediate. In this case we are done. So we may assume that ${\varSigma}^{\ast}\models {\mathrm{aBx}}_2$ . Then ${\varSigma}^{\ast}\models \exists {\mathrm{x}}_3{\mathrm{x}}_2={\mathrm{ax}}_3$ , that is, ${\varSigma}^{\ast}\models {\mathrm{b}}^{\mathrm{n}}{\mathrm{ay}}_1\dots {\mathrm{y}}_{\mathrm{n}}=\mathrm{u}={\mathrm{b}}^{\mathrm{n}}{\mathrm{ax}}_3\mathrm{aa}$ . Thus ${\varSigma}^{\ast}\models {\mathrm{x}}_3\mathrm{aaEu}$ , and from , by 2.3(b), we have that ${\varSigma}^{\ast}\models \alpha ({\mathrm{x}}_3\mathrm{aa})\ge \beta ({\mathrm{x}}_3\mathrm{aa})+\underline{1}$ . Then ${\mathrm{x}}_3\mathrm{aa}$ has at least one initial segment $\mathrm{v}$ , namely itself, such that ${\varSigma}^{\ast}\models \alpha (\mathrm{v})\ge \beta (\mathrm{v})+\underline{1}$ . Let ${\mathrm{u}}_1$ be the shortest such initial segment of ${\mathrm{x}}_3\mathrm{aa}$ .

By choice of ${\mathrm{u}}_1$ we have that ${\varSigma}^{\ast}\models \alpha ({\mathrm{u}}_1)\ge \beta ({\mathrm{u}}_1)+1$ . We may assume that ${\varSigma}^{\ast}\models {\mathrm{u}}_1\ne \mathrm{a}$ ; otherwise, we are done. Suppose, for a reductio, that ${\varSigma}^{\ast}\models \alpha ({\mathrm{u}}_1)>\beta ({\mathrm{u}}_1)+\underline{1}$ . Then ${\varSigma}^{\ast}\models {\mathrm{u}}_1\ne \mathrm{b}$ .

Suppose that ${\varSigma}^{\ast}\models {\mathrm{aEu}}_1$ . Then ${\varSigma}^{\ast}\models \exists \mathrm{z}\;{\mathrm{u}}_1=\mathrm{za}$ . Then from hypothesis ${\varSigma}^{\ast}\models \alpha ({\mathrm{u}}_1)>\beta ({\mathrm{u}}_1)+\underline{1}$ , we obtain ${\varSigma}^{\ast}\models \alpha (\mathrm{z})=\alpha ({\mathrm{u}}_1)\hbox{--} \underline{1}>\beta ({\mathrm{u}}_1)=\beta (\mathrm{z})$ , so ${\varSigma}^{\ast}\models \alpha (\mathrm{z})\ge \beta (\mathrm{z})+\underline{1}$ . Since ${\varSigma}^{\ast}\models {\mathrm{zBu}}_1\ \&\ ({\mathrm{u}}_1{\mathrm{Bx}}_3\mathrm{aa}\;\mathrm{v}\;{\mathrm{u}}_1={\mathrm{x}}_3\mathrm{aa})$ , we have that ${\varSigma}^{\ast}\models {\mathrm{zBx}}_3\mathrm{aa}$ . But this contradicts the choice of ${\mathrm{u}}_1$ . Hence ${\varSigma}^{\ast}\models \neg {\mathrm{aEu}}_1$ .

Suppose that ${\varSigma}^{\ast}\models {\mathrm{bEu}}_1$ . Then ${\varSigma}^{\ast}\models \exists \mathrm{z}\;{\mathrm{u}}_1=\mathrm{zb}$ ; hence

$$\begin{align*}{\varSigma}^{\ast}\models \alpha (\mathrm{z})=\alpha (\mathrm{z}\mathrm{b})=\alpha ({\mathrm{u}}_1)\ge \beta ({\mathrm{u}}_1)+\underline{1}=\beta (\mathrm{z}\mathrm{b})+\underline{1}>\beta (\mathrm{z})+\underline{1},\end{align*}$$

where ${\varSigma}^{\ast}\models {\mathrm{zBu}}_1\ \&\ ({\mathrm{u}}_1{\mathrm{Bx}}_3\mathrm{aa}\;\mathrm{v}\;{\mathrm{u}}_1={\mathrm{x}}_3\mathrm{aa})$ . So ${\varSigma}^{\ast}\models {\mathrm{zBx}}_3\mathrm{aa}$ , again contradicting the choice of ${\mathrm{u}}_1$ . Hence also ${\varSigma}^{\ast}\models \neg {\mathrm{bEu}}_1$ .

But then we have that ${\varSigma}^{\ast}\models {\mathrm{u}}_1\ne \mathrm{a}\ \&\ {\mathrm{u}}_1\ne \mathrm{b}\ \&\ \neg {\mathrm{aEu}}_1\ \&\ \neg {\mathrm{bEu}}_1$ , which is impossible. Therefore ${\varSigma}^{\ast}\models \alpha ({\mathrm{u}}_1)=\beta ({\mathrm{u}}_1)+\underline{1}$ . By choice of ${\mathrm{u}}_1$ it follows also that ${\varSigma}^{\ast}\models \forall \mathrm{w}\;({\mathrm{wBu}}_1\to \alpha (\mathrm{w})\le \beta (\mathrm{w}))$ . This completes the proof of Claim 1.

We have that ${\varSigma}^{\ast}\models {\mathrm{u}}_1{\mathrm{Bx}}_3\mathrm{aa}\;\mathrm{v}\;{\mathrm{u}}_1={\mathrm{x}}_3\mathrm{aa}$ .

If ${\varSigma}^{\ast}\models {\mathrm{u}}_1={\mathrm{x}}_3\mathrm{aa}$ , then ${\varSigma}^{\ast}\models {\mathrm{b}}^{\mathrm{n}}{\mathrm{au}}_1={\mathrm{b}}^{\mathrm{n}}{\mathrm{ax}}_3\mathrm{aa}=\mathrm{u}$ . Since , we then have ${\varSigma}^{\ast}\models \alpha (\mathrm{u})=\alpha ({\mathrm{b}}^{\mathrm{n}}{\mathrm{au}}_1)=\underline{1}+\alpha ({\mathrm{u}}_1)=\underline{1}+(\beta ({\mathrm{u}}_1)+\underline{1})=\beta ({\mathrm{u}}_1)+\underline{2}$ , whereas ${\varSigma}^{\ast}\models \beta (\mathrm{u})=\beta ({\mathrm{b}}^{\mathrm{n}}{\mathrm{au}}_1)=\underline {\mathrm{n}}+\beta ({\mathrm{u}}_1)$ . But from hypothesis we have ${\varSigma}^{\ast}\models \alpha (\mathrm{u})=\beta (\mathrm{u})+1$ . Hence ${\varSigma}^{\ast}\models \beta ({\mathrm{u}}_1)+\underline{2}=\beta ({\mathrm{u}}_1)+\underline {\mathrm{n}}+\underline{1}$ , contradicting hypothesis $\mathrm{n}\ge 2$ . Therefore ${\varSigma}^{\ast}\models {\mathrm{u}}_1\ne {\mathrm{x}}_3\mathrm{aa}$ . But then it follows that ${\varSigma}^{\ast}\models {\mathrm{u}}_1{\mathrm{Bx}}_3\mathrm{aa}$ . So ${\varSigma}^{\ast}\models \exists {\mathrm{v}}_1{\mathrm{x}}_3\mathrm{aa}={\mathrm{u}}_1{\mathrm{v}}_1$ , whence ${\varSigma}^{\ast}\models {\mathrm{b}}^{\mathrm{n}}{\mathrm{au}}_1{\mathrm{v}}_1={\mathrm{b}}^{\mathrm{n}}{\mathrm{ax}}_3\mathrm{aa}=\mathrm{u}$ , which means that ${\varSigma}^{\ast}\models {\mathrm{b}}^{\mathrm{n}}{\mathrm{au}}_1\mathrm{Bu}$ .

Claim 2. For each $\mathrm{i}$ , $1\le \mathrm{i}<\mathrm{n}$ ,

Assume that ${\mathrm{v}}_1,\dots, {\mathrm{v}}_{\mathrm{i}}$ have been picked such that

Then ${\varSigma}^{\ast}\models \exists {\mathrm{w}}_{\mathrm{i}}\;\mathrm{u}={\mathrm{b}}^{\mathrm{n}}{\mathrm{av}}_1\dots {\mathrm{v}}_{\mathrm{i}}{\mathrm{w}}_{\mathrm{i}}$ ; hence ${\varSigma}^{\ast}\models {\mathrm{w}}_{\mathrm{i}}\mathrm{Eu}$ . From hypothesis by 2.3(b) we then have that ${\varSigma}^{\ast}\models \alpha ({\mathrm{w}}_{\mathrm{i}})\ge \beta ({\mathrm{w}}_{\mathrm{i}})+\underline{1}$ . Then we have

$$\begin{align*} \begin{array}{l}\hspace{-5pt}{\varSigma}^{\ast}\models \alpha (\mathrm{u})=\alpha ({\mathrm{b}}^{\mathrm{n}}{\mathrm{av}}_1\dots {\mathrm{v}}_{\mathrm{i}}{\mathrm{w}}_{\mathrm{i}})=\underline{1}+\alpha ({\mathrm{v}}_1\dots {\mathrm{v}}_{\mathrm{i}})+\alpha ({\mathrm{w}}_{\mathrm{i}})=\\ {}\hspace{40pt}=\underline{1}+{\sum}_{1\le \mathrm{j}\le \mathrm{i}}\alpha ({\mathrm{v}}_{\mathrm{j}})+\alpha ({\mathrm{w}}_{\mathrm{i}})=\underline{1}+{\sum}_{1\le \mathrm{j}\le \mathrm{i}}(\beta ({\mathrm{v}}_{\mathrm{j}})+\underline{1})+\alpha ({\mathrm{w}}_{\mathrm{i}})=\\ \hspace{40pt} =\underline{1}+\underline {\mathrm{i}}+{\sum}_{1\le \mathrm{j}\le \mathrm{i}}\beta ({\mathrm{v}}_{\mathrm{j}})+\alpha ({\mathrm{w}}_{\mathrm{i}}),\end{array}\end{align*}$$

whereas ${\begin{array}{l}{\varSigma}^{\ast}\models \beta (\mathrm{u})=\beta ({\mathrm{b}}^{\mathrm{n}}{\mathrm{av}}_1\dots {\mathrm{v}}_{\mathrm{i}}{\mathrm{w}}_{\mathrm{i}})=\underline {\mathrm{n}}+\beta ({\mathrm{v}}_1\dots {\mathrm{v}}_{\mathrm{i}})+\beta ({\mathrm{w}}_{\mathrm{i}})=\\ \quad\qquad\qquad\qquad\qquad\qquad\hspace{20pt} =\underline {\mathrm{n}}+{\sum}_{1\le \mathrm{j}\le \mathrm{i}}\beta ({\mathrm{v}}_{\mathrm{j}})+\beta ({\mathrm{w}}_{\mathrm{i}}).\end{array}}$

But from hypothesis we have ${\varSigma}^{\ast}\models \alpha (\mathrm{u})=\beta (\mathrm{u})+\underline{1}$ . Then ${\varSigma}^{\ast}\models \underline{1}+\underline {\mathrm{i}}+{\sum}_{1\le \mathrm{j}\le \mathrm{i}}\beta ({\mathrm{v}}_{\mathrm{j}})+\alpha ({\mathrm{w}}_{\mathrm{i}})=\underline {\mathrm{n}}+{\sum}_{1\le \mathrm{j}\le \mathrm{i}}\beta ({\mathrm{v}}_{\mathrm{j}})+\beta ({\mathrm{w}}_{\mathrm{i}})+\underline{1}$ , whence

(#) $$\begin{align}\qquad\qquad\quad{\varSigma}^{\ast}\models \underline {\mathrm{i}}+\alpha ({\mathrm{w}}_{\mathrm{i}})=\underline {\mathrm{n}}+\beta ({\mathrm{w}}_{\mathrm{i}}).\end{align}$$

We pick ${\mathrm{v}}_{\mathrm{i}+1}$ as follows: if ${\varSigma}^{\ast}\models {\mathrm{w}}_{\mathrm{i}}=\mathrm{a}$ , we let ${\mathrm{v}}_{\mathrm{i}+1}=\mathrm{a}$ ; if ${\varSigma}^{\ast}\models {\mathrm{w}}_{\mathrm{i}}\ne \mathrm{a}$ , let ${\mathrm{v}}_{\mathrm{i}+1}$ be the shortest initial segment $\mathrm{v}$ of ${\mathrm{w}}_{\mathrm{i}}$ such that ${\varSigma}^{\ast}\models \alpha (\mathrm{v})\ge \beta (\mathrm{v})+\underline{1}$ .

If ${\varSigma}^{\ast}\models {\mathrm{w}}_{\mathrm{i}}=\mathrm{a}$ , then ${\varSigma}^{\ast}\models \mathrm{u}={\mathrm{b}}^{\mathrm{n}}{\mathrm{av}}_1\dots {\mathrm{v}}_{\mathrm{i}}{\mathrm{w}}_{\mathrm{i}}={\mathrm{b}}^{\mathrm{n}}{\mathrm{av}}_1\dots {\mathrm{v}}_{\mathrm{i}}\mathrm{a}$ , and we of course have that . Since ${\varSigma}^{\ast}\models {\mathrm{w}}_{\mathrm{i}}=\mathrm{a}$ , we have ${\varSigma}^{\ast}\models \alpha ({\mathrm{w}}_{\mathrm{i}})=\underline{1}\ \&\ \beta ({\mathrm{w}}_{\mathrm{i}})=\underline{0}$ , so from $(\#)$ we obtain ${\varSigma}^{\ast}\models \underline {\mathrm{i}}+\underline{1}=\underline {\mathrm{n}}$ .

If ${\varSigma}^{\ast}\models {\mathrm{w}}_{\mathrm{i}}\ne \mathrm{a}$ , then the same argument as in Claim 1 with ${\mathrm{v}}_{\mathrm{i}+1}$ in place of ${\mathrm{u}}_1$ and ${\mathrm{w}}_{\mathrm{i}}$ in place of ${\mathrm{x}}_3\mathrm{aa}$ shows that ${\varSigma}^{\ast}\models \alpha ({\mathrm{v}}_{\mathrm{i}+1})=\beta ({\mathrm{v}}_{\mathrm{i}+1})+\underline{1}$ , and that . By choice of ${\mathrm{v}}_{\mathrm{i}+1}$ we have that ${\varSigma}^{\ast}\models {\mathrm{v}}_{\mathrm{i}+1}={\mathrm{w}}_{\mathrm{i}}\;\mathrm{v}\;{\mathrm{v}}_{\mathrm{i}+1}{\mathrm{Bw}}_{\mathrm{i}}$ .

If ${\varSigma}^{\ast}\models {\mathrm{v}}_{\mathrm{i}+1}={\mathrm{w}}_{\mathrm{i}}$ , then ${\varSigma}^{\ast}\models \mathrm{u}={\mathrm{b}}^{\mathrm{n}}{\mathrm{av}}_1\dots {\mathrm{v}}_{\mathrm{i}}{\mathrm{w}}_{\mathrm{i}}={\mathrm{b}}^{\mathrm{n}}{\mathrm{av}}_1\dots {\mathrm{v}}_{\mathrm{i}}{\mathrm{v}}_{\mathrm{i}+1}$ . Then ${\varSigma}^{\ast}\models \alpha ({\mathrm{w}}_{\mathrm{i}})=\beta ({\mathrm{w}}_{\mathrm{i}})+\underline{1}$ , so from $(\#)$ we obtain

${\varSigma}^{\ast}\models \underline {\mathrm{i}}+\beta ({\mathrm{w}}_{\mathrm{i}})+\underline{1}=\underline {\mathrm{n}}+\beta ({\mathrm{w}}_{\mathrm{i}})$ , whence ${\varSigma}^{\ast}\models \underline {\mathrm{i}}+\underline{1}=\underline {\mathrm{n}}$ .

If ${\varSigma}^{\ast}\models {\mathrm{v}}_{\mathrm{i}+1}{\mathrm{Bw}}_{\mathrm{i}}$ , then ${\varSigma}^{\ast}\models \exists {\mathrm{w}}_{\mathrm{i}+1}{\mathrm{w}}_{\mathrm{i}}={\mathrm{v}}_{\mathrm{i}+1}{\mathrm{w}}_{\mathrm{i}+1}$ . So

$$\begin{align*}{\varSigma}^{\ast}\models \mathrm{u}={\mathrm{b}}^{\mathrm{n}}{\mathrm{av}}_1\dots {\mathrm{v}}_{\mathrm{i}}{\mathrm{w}}_{\mathrm{i}}={\mathrm{b}}^{\mathrm{n}}{\mathrm{av}}_1\dots {\mathrm{v}}_{\mathrm{i}}({\mathrm{v}}_{\mathrm{i}+1}{\mathrm{w}}_{\mathrm{i}+1}).\end{align*}$$

Then we have that

$$\begin{align*}{\varSigma}^{\ast}\models \alpha ({\mathrm{w}}_{\mathrm{i}})=\alpha ({\mathrm{v}}_{\mathrm{i}+1}{\mathrm{w}}_{\mathrm{i}+1})=\alpha ({\mathrm{v}}_{\mathrm{i}+1})+\alpha ({\mathrm{w}}_{\mathrm{i}+1})=\beta ({\mathrm{v}}_{\mathrm{i}+1})+\underline{1}+\alpha ({\mathrm{w}}_{\mathrm{i}+1})\end{align*}$$

and $\beta ({\mathrm{w}}_{\mathrm{i}})=\beta ({\mathrm{v}}_{\mathrm{i}+1}{\mathrm{w}}_{\mathrm{i}+1})=\beta ({\mathrm{v}}_{\mathrm{i}+1})+\beta ({\mathrm{w}}_{\mathrm{i}+1})$ . Hence from $(\#)$ we obtain ${\varSigma}^{\ast}\models \underline {\mathrm{i}}+\beta ({\mathrm{v}}_{\mathrm{i}+1})+\underline{1}+\alpha ({\mathrm{w}}_{\mathrm{i}+1})=\underline {\mathrm{n}}+\beta ({\mathrm{v}}_{\mathrm{i}+1})+\beta ({\mathrm{w}}_{\mathrm{i}+1})$ . But ${\varSigma}^{\ast}\models {\mathrm{w}}_{\mathrm{i}+1}\mathrm{Eu}$ , so from by 2.3(b), we have ${\varSigma}^{\ast}\models \alpha ({\mathrm{w}}_{\mathrm{i}+1})\ge \beta ({\mathrm{w}}_{\mathrm{i}+1})+\underline{1}>\beta ({\mathrm{w}}_{\mathrm{i}+1})$ . But then ${\varSigma}^{\ast}\models \underline {\mathrm{i}}+\underline{1}<\mathrm{n}\ \&\ ({\mathrm{b}}^{\mathrm{n}}{\mathrm{av}}_1\dots {\mathrm{v}}_{\mathrm{i}}{\mathrm{v}}_{\mathrm{i}+1})\mathrm{Bu}$ .

This completes the proof of Claim 2.

From Claim 1 and $\mathrm{n}\hbox{--} 1$ applications of Claim 2 we obtain strings ${\mathrm{u}}_1,\dots, {\mathrm{u}}_{\mathrm{n}}$ such that . Hence ${\varSigma}^{\ast}\models {\mathrm{y}}_1\dots {\mathrm{y}}_{\mathrm{n}}={\mathrm{u}}_1\dots {\mathrm{u}}_{\mathrm{n}}$ . Then uniqueness follows from 2.4.

Proof Assume . Then . We proceed exactly as in the proof of Theorem 2.11. If ${\varSigma}^{\ast}\models \mathrm{u}={\mathrm{b}}^{\mathrm{n}}{\mathrm{ay}}_1\dots {\mathrm{y}}_{\mathrm{n}}$ we have that $\mathrm{n}\le \mathrm{m}$ . If $\mathrm{n}=1$ and ${\varSigma}^{\ast}\models {\mathrm{u}}_1=\mathrm{a}$ , then , and if $\mathrm{n}=2$ and ${\varSigma}^{\ast}\models {\mathrm{u}}_1={\mathrm{u}}_2=\mathrm{a}$ , then and so . In the general case, once we reach the end of the proof of Theorem 2.11, since ${\varSigma}^{\ast}\models {\mathrm{y}}_1\dots {\mathrm{y}}_{\mathrm{n}}={\mathrm{u}}_1\dots {\mathrm{u}}_{\mathrm{n}}$ we have that ${\varSigma}^{\ast }{\models }\wedge_{1\le \mathrm{i}\le \mathrm{n}}{\mathrm{u}}_{\mathrm{i}}{\subseteq}_{\mathrm{p}}\mathrm{u}$ . But then from the principal hypothesis it follows that as claimed.

Remark. If we let

Proof Assume . Then , and by 2.6 we have that, for some $\mathrm{n}\ge 1$ , ${\varSigma}^{\ast}\models \exists {\mathrm{u}}_1\dots {\mathrm{u}}_{\mathrm{n}}\mathrm{u}={\mathrm{b}}^{\mathrm{n}}{\mathrm{au}}_1\dots {\mathrm{u}}_{\mathrm{n}}$ . Then ${\varSigma}^{\ast}\models {\mathrm{Tally}}_{\mathrm{b}}({\mathrm{b}}^{\mathrm{n}})\ \&\ {\mathrm{b}}^{\mathrm{n}}\mathrm{aBu}$ , so from we have ${\varSigma}^{\ast}\models {\mathrm{b}}^{\mathrm{n}}={\mathrm{b}}^{\mathrm{m}}$ , whence $\mathrm{n}=\mathrm{m}$ . But then we may take ${\mathrm{x}}_1,\dots, {\mathrm{x}}_{\mathrm{m}}$ to be ${\mathrm{u}}_1,\dots, {\mathrm{u}}_{\mathrm{n}}$ .

Proof We rely on the coding scheme for sequences of strings explained in [Reference Damnjanovic3, Section 4], and follow the notation used there. In particular, we have that ${\varSigma}^{\ast}\models \mathrm{Pair}[\mathrm{x},\mathrm{y},\mathrm{z}]$ iff string $\mathrm{z}$ codes the pair of strings $\mathrm{x}$ and $\mathrm{y}$ , ${\varSigma}^{\ast}\models \mathrm{Set}(\mathrm{x})$ iff $\mathrm{x}$ is a set code, and ${\varSigma}^{\ast}\models \mathrm{x}\;\varepsilon\;\mathrm{y}$ just in case string $\mathrm{x}$ is a member of the set coded by string $\mathrm{y}$ . Let ${\mathrm{R}}_{\mathrm{k}}{(}\mathrm{u},\mathrm{x})$ abbreviate the ${{\mathcal{L}}}_{\mathrm{C}}$ -formula

Let ${\mathrm{Comp}}_{\le \mathrm{n}}{(}\mathrm{u},\mathrm{x})$ abbreviate

Let ${\mathrm{MinComp}}_{\le \mathrm{n}}{(}\mathrm{u},\mathrm{x})$ abbreviate

We then let $\mathrm{F}(\mathrm{x},\mathrm{y})\equiv \exists \mathrm{u},\mathrm{w}({\mathrm{MinComp}}_{\le \mathrm{n}}(\mathrm{u},\mathrm{x})\ \&\ \mathrm{Pair}[\mathrm{x},\mathrm{y},\mathrm{w}]\ \&\ \mathrm{w}\;\varepsilon\;\mathrm{u}).$

Now, suppose (a) and (b) hold. For (i), assume that . We argue by induction on the generating relation of strings. If ${\varSigma}^{\ast}\models \mathrm{x}=\mathrm{a}$ , let ${\mathrm{w}}_0$ be the string such that ${\varSigma}^{\ast}\models \mathrm{Pair}[\mathrm{a},\mathrm{p},{\mathrm{w}}_0]$ , and let ${\mathrm{u}}_0$ be the string that codes the singleton sequence of ${\mathrm{w}}_0$ . Then the desired claim follows immediately from (a) and the definition of $\mathrm{F}(\mathrm{x},\mathrm{y})$ . Assume ${\varSigma}^{\ast}\models \mathrm{x}={\mathrm{b}}^{\mathrm{k}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{k}}$ where the claim holds for ${\mathrm{x}}_1,\dots, {\mathrm{x}}_{\mathrm{k}}$ with $\mathrm{k}\le \mathrm{n}$ . Then

Let ${\mathrm{w}}_1,\dots, {\mathrm{w}}_{\mathrm{k}}$ be the strings such that ${\varSigma}^{\ast }{\models }\wedge_{1\le \mathrm{i}\le \mathrm{k}}\mathrm{Pair}[{\mathrm{x}}_{\mathrm{i}},{\mathrm{y}}_{\mathrm{i}},{\mathrm{w}}_{\mathrm{i}}]$ , and let $\mathrm{u}$ be the string that codes the sequence ${\mathrm{w}}_1,\dots, {\mathrm{w}}_{\mathrm{k}}$ . From (b) we have that . Pick the string $\mathrm{w}$ such that ${\varSigma}^{\ast}\models \mathrm{Pair}[{\mathrm{b}}^{\mathrm{k}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{k}},\mathrm{y},\mathrm{w}]$ , and let $\mathrm{u}'$ be the string that codes the sequence ${\mathrm{w}}_1,\dots, {\mathrm{w}}_{\mathrm{k}},\mathrm{w}$ . Then the claim follows immediately from the definition of $\mathrm{F}(\mathrm{x},\mathrm{y})$ . (ii) and (iii) follow straightforwardly from the definition of $\mathrm{F}(\mathrm{x},\mathrm{y})$ and (i).

Proof (a) Assume $\mathrm{M}\models {({\mathrm{T}}_{\mathrm{n}}({\mathrm{y}}_1,\dots, {\mathrm{y}}_{\mathrm{n}},0))}^{\tau }$ where $\mathrm{M}{\models }\wedge_{1\le \mathrm{i}\le \mathrm{n}}\mathrm{I}({\mathrm{y}}_{\mathrm{i}})$ . Then

$$\begin{align*}\mathrm{M}\models \exists {\mathrm{u}}_1\dots {\mathrm{u}}_{\mathrm{n}}({\mathrm{u}}_1={\tau}_2({\mathrm{y}}_{\mathrm{n}},0)\ \&\ {\wedge}_{1\le \mathrm{i}<\mathrm{n}}{\mathrm{u}}_{\mathrm{i}+1}={\tau}_2({\mathrm{y}}_{\mathrm{n}-\mathrm{i}},{\mathrm{u}}_{\mathrm{i}})\ \&\ {\tau}_1(0)={\tau}_1({\mathrm{u}}_{\mathrm{n}})).\end{align*}$$

Suppose $\mathrm{n}>1$ . Then $\mathrm{M}\models {\tau}_1(0)={\tau}_1({\mathrm{u}}_{\mathrm{n}})\ \&\ {\mathrm{u}}_{\mathrm{n}}={\tau}_2({\mathrm{y}}_1,{\mathrm{u}}_{\mathrm{n}-1})$ . But then, by $(\mathrm{T}{3}_1)$ , we have $\mathrm{M}\models 0={\tau}_2({\mathrm{y}}_1,{\mathrm{u}}_{\mathrm{n}-1})$ , contradicting $(\mathrm{T}{1}_2)$ . If $\mathrm{n}=1$ , then $\mathrm{M}\models \exists {\mathrm{u}}_1({\mathrm{u}}_1={\tau}_2({\mathrm{y}}_1,0)\ \&\ {\tau}_1(0)={\tau}_1({\mathrm{u}}_1))$ . Then, by $(\mathrm{T}{3}_1)$ , $\mathrm{M}\models 0={\tau}_2({\mathrm{y}}_1,0)$ , contradicting $(\mathrm{T}{1}_2)$ . Hence $\mathrm{M}\models \neg {({\mathrm{T}}_{\mathrm{n}}({\mathrm{y}}_1,\dots, {\mathrm{y}}_{\mathrm{n}},0))}^{\tau }$ .

(b) Assume $\mathrm{M}\models {({\mathrm{T}}_{\mathrm{k}}({\mathrm{x}}_1,\dots, {\mathrm{x}}_{\mathrm{k}},\mathrm{x}))}^{\tau}\ \&\ {({\mathrm{T}}_{\mathrm{n}}({\mathrm{y}}_1,\dots, {\mathrm{y}}_{\mathrm{n}},\mathrm{y}))}^{\tau}\ \&\ \mathrm{x}=\mathrm{y}$ where $\mathrm{M}{\models }\wedge_{1\le \mathrm{i}\le \mathrm{k}}\mathrm{I}({\mathrm{x}}_{\mathrm{i}})\ \&\ {\wedge}_{1\le \mathrm{j}\le \mathrm{n}}\mathrm{I}({\mathrm{y}}_{\mathrm{j}})\ \&\ \mathrm{I}(\mathrm{x})\ \&\ \mathrm{I}(\mathrm{y})$ . Then $\mathrm{M}\models \exists {\mathrm{u}}_1\dots {\mathrm{u}}_{\mathrm{k}}({\mathrm{u}}_1={\tau}_2({\mathrm{x}}_{\mathrm{k}},0)\ \&\ {\wedge}_{1\le \mathrm{i}<\mathrm{k}}{\mathrm{u}}_{\mathrm{i}+1}={\tau}_2({\mathrm{x}}_{\mathrm{k}-\mathrm{i}},{\mathrm{u}}_{\mathrm{i}})\ \&\ \mathrm{x}={\tau}_1({\mathrm{u}}_{\mathrm{k}}))$ and $\mathrm{M}\models \exists {\mathrm{v}}_1\dots {\mathrm{v}}_{\mathrm{n}}({\mathrm{v}}_1={\tau}_2({\mathrm{y}}_{\mathrm{n}},0)\ \&\ {\wedge}_{1\le \mathrm{j}<\mathrm{n}}{\mathrm{v}}_{\mathrm{j}+1}={\tau}_2({\mathrm{y}}_{\mathrm{n}-\mathrm{j}},{\mathrm{v}}_{\mathrm{j}})\ \&\ \mathrm{y}={\tau}_1({\mathrm{v}}_{\mathrm{n}}))$ , whence $\mathrm{M}\models {\tau}_1({\mathrm{u}}_{\mathrm{k}})=\mathrm{x}=\mathrm{y}={\tau}_1({\mathrm{v}}_{\mathrm{n}})$ , and further, by $(\mathrm{T}{3}_1)$ , $\mathrm{M}\models {\mathrm{u}}_{\mathrm{k}}={\mathrm{v}}_{\mathrm{n}}$ .

Let ${\mathrm{x}}_0=\mathrm{x}$ and ${\mathrm{y}}_0=\mathrm{y}$ . By hypothesis, we have that

(1) $$\begin{align}\mathrm{M}\models {\mathrm{x}}_0={\mathrm{y}}_0.\end{align}$$

Assume now that $\mathrm{M}\models {\mathrm{x}}_{\mathrm{j}}={\mathrm{y}}_{\mathrm{j}}\ \&\ {\mathrm{u}}_{\mathrm{k}-\mathrm{j}}={\mathrm{v}}_{\mathrm{n}-\mathrm{j}}$ , where $0\le \mathrm{j}<\mathrm{k}$ . Then $\mathrm{M}\models {\tau}_2({\mathrm{x}}_{\mathrm{j}+1},{\mathrm{u}}_{\mathrm{k}-(\mathrm{j}+1)})={\mathrm{u}}_{\mathrm{k}-\mathrm{j}}={\mathrm{v}}_{\mathrm{n}-\mathrm{j}}={\tau}_2({\mathrm{y}}_{\mathrm{j}+1},{\mathrm{v}}_{\mathrm{n}-(\mathrm{j}+1)})$ , whence, by $(\mathrm{T}{3}_2)$ , $\mathrm{M}\models {\mathrm{x}}_{\mathrm{j}+1}={\mathrm{y}}_{\mathrm{j}+1}\ \&\ {\mathrm{u}}_{\mathrm{k}-(\mathrm{j}+1)}={\mathrm{v}}_{\mathrm{n}-(\mathrm{j}+1)}$ . Thus, we have

(2) $$\begin{align}\mathrm{M}\models {\mathrm{x}}_{\mathrm{j}}={\mathrm{y}}_{\mathrm{j}}\ \&\ {\mathrm{u}}_{\mathrm{k}-\mathrm{j}}={\mathrm{v}}_{\mathrm{n}-\mathrm{j}}\to {\mathrm{x}}_{\mathrm{j}+1}={\mathrm{y}}_{\mathrm{j}+1}\ \&\ {\mathrm{u}}_{\mathrm{k}-(\mathrm{j}+1)}={\mathrm{v}}_{\mathrm{n}-(\mathrm{j}+1)}.\end{align}$$

From (1) after $\mathrm{k}$ applications of (2) we obtain

$$\begin{align*}\mathrm{M}\models {\mathrm{x}}_{\mathrm{k}}={\mathrm{y}}_{\mathrm{k}}\ \&\ {\mathrm{u}}_1={\mathrm{u}}_{\mathrm{k}-(\mathrm{k}-1)}={\mathrm{v}}_{\mathrm{n}-(\mathrm{k}-1)}={\mathrm{v}}_{\mathrm{n}-\mathrm{k}+1}.\end{align*}$$

But $\mathrm{M}\models {\mathrm{u}}_1={\tau}_2({\mathrm{x}}_{\mathrm{k}},0)$ and $\mathrm{M}\models {\mathrm{v}}_{\mathrm{n}-\mathrm{k}+1}={\tau}_2({\mathrm{y}}_{\mathrm{k}},{\mathrm{v}}_{\mathrm{n}-\mathrm{k}})$ where $\mathrm{n}\hbox{--} \mathrm{k}\ne 0$ . Hence $\mathrm{M}\models {\tau}_2({\mathrm{x}}_{\mathrm{k}},0)={\tau}_2({\mathrm{y}}_{\mathrm{k}},{\mathrm{v}}_{\mathrm{n}-\mathrm{k}})$ . But then, by $(\mathrm{T}{3}_2)$ , $\mathrm{M}\models 0={\mathrm{v}}_{\mathrm{n}-\mathrm{k}}={\tau}_2({\mathrm{y}}_{\mathrm{k}+1},{\mathrm{v}}_{\mathrm{n}-(\mathrm{k}+1)})$ , contradicting $(\mathrm{T}{1}_2)$ . Hence $\mathrm{M}\models \mathrm{x}\ne \mathrm{y}$ , and we obtain $\mathrm{M}\models {(\mathrm{T}{2}_{\mathrm{k},\mathrm{n}})}^{\tau }$ .

(c) Assume $\mathrm{M}\models {({\mathrm{T}}_{\mathrm{n}}({\mathrm{x}}_1,\dots, {\mathrm{x}}_{\mathrm{n}},\mathrm{x}))}^{\tau}\ \&\ {({\mathrm{T}}_{\mathrm{n}}({\mathrm{y}}_1,\dots, {\mathrm{y}}_{\mathrm{n}},\mathrm{y}))}^{\tau}\ \&\ \mathrm{x}=\mathrm{y}$ where $\mathrm{M}{\models }\wedge_{1\le \mathrm{i}\le \mathrm{n}}\mathrm{I}({\mathrm{x}}_{\mathrm{i}})\ \&\ {\wedge}_{1\le \mathrm{j}\le \mathrm{n}}\mathrm{I}({\mathrm{y}}_{\mathrm{j}})\ \&\ \mathrm{I}(\mathrm{x})\ \&\ \mathrm{I}(\mathrm{y})$ . Then $\mathrm{M}\models \exists {\mathrm{u}}_1\dots {\mathrm{u}}_{\mathrm{k}}({\mathrm{u}}_1={\tau}_2({\mathrm{x}}_{\mathrm{n}},0)\ \&\ {\wedge}_{1\le \mathrm{i}<\mathrm{n}}{\mathrm{u}}_{\mathrm{i}+1}={\tau}_2({\mathrm{x}}_{\mathrm{n}-\mathrm{i}},{\mathrm{u}}_{\mathrm{i}})\ \&\ \mathrm{x}={\tau}_1({\mathrm{u}}_{\mathrm{n}}))$ and $\mathrm{M}\models \exists {\mathrm{v}}_1\dots {\mathrm{v}}_{\mathrm{n}}({\mathrm{v}}_1={\tau}_2({\mathrm{y}}_{\mathrm{n}},0)\ \&\ {\wedge}_{1\le \mathrm{j}<\mathrm{n}}{\mathrm{v}}_{\mathrm{j}+1}={\tau}_2({\mathrm{y}}_{\mathrm{n}-\mathrm{j}},{\mathrm{v}}_{\mathrm{j}})\ \&\ \mathrm{y}={\tau}_1({\mathrm{v}}_{\mathrm{n}}))$ , whence $\mathrm{M}\models {\tau}_1({\mathrm{u}}_{\mathrm{n}})=\mathrm{x}=\mathrm{y}={\tau}_1({\mathrm{v}}_{\mathrm{n}})$ , and further, by $(\mathrm{T}{3}_1)$ , $\mathrm{M}\models {\mathrm{u}}_{\mathrm{n}}={\mathrm{v}}_{\mathrm{n}}$ . We now show, for $0\le \mathrm{j}<\mathrm{n}$ , that

(*) $$\begin{align}\kern0.36em \mathrm{M}\models {\mathrm{x}}_{\mathrm{j}}={\mathrm{y}}_{\mathrm{j}}\ \&\ {\mathrm{u}}_{\mathrm{n}-\mathrm{j}}={\mathrm{v}}_{\mathrm{n}-\mathrm{j}}\to {\mathrm{x}}_{\mathrm{j}+1}={\mathrm{y}}_{\mathrm{j}+1}\ \&\ {\mathrm{u}}_{\mathrm{n}-(\mathrm{j}+1)}={\mathrm{v}}_{\mathrm{n}-(\mathrm{j}+1)}.\end{align}$$

Assume $\mathrm{M}\models {\mathrm{x}}_{\mathrm{j}}={\mathrm{y}}_{\mathrm{j}}\ \&\ {\mathrm{u}}_{\mathrm{n}-\mathrm{j}}={\mathrm{v}}_{\mathrm{n}-\mathrm{j}}$ . Then $\mathrm{M}\models {\tau}_2({\mathrm{x}}_{\mathrm{j}+1},{\mathrm{u}}_{\mathrm{n}-(\mathrm{j}+1)})={\mathrm{u}}_{\mathrm{n}-\mathrm{j}}={\mathrm{v}}_{\mathrm{n}-\mathrm{j}}={\tau}_2({\mathrm{y}}_{\mathrm{j}+1},{\mathrm{v}}_{\mathrm{n}-(\mathrm{j}+1)})$ , whence, by $(\mathrm{T}{3}_2)$ , $\mathrm{M}\models {\mathrm{x}}_{\mathrm{j}+1}={\mathrm{y}}_{\mathrm{j}+1}\ \&\ {\mathrm{u}}_{\mathrm{n}-(\mathrm{j}+1)}={\mathrm{v}}_{\mathrm{n}-(\mathrm{j}+1)}$ _, as claimed.

Letting ${\mathrm{x}}_0=\mathrm{x}$ and ${\mathrm{y}}_0=\mathrm{y}$ , after $\mathrm{n}$ applications of $({}^\ast )$ we obtain $\mathrm{M}\models {\mathrm{x}}_{\mathrm{n}}={\mathrm{y}}_{\mathrm{n}}$ . Hence $\mathrm{M}{\models }\wedge_{1\le \mathrm{i}\le \mathrm{n}}{\mathrm{x}}_{\mathrm{i}}={\mathrm{y}}_{\mathrm{i}}$ as needed.

(d) Assume $\mathrm{M}\models {({\mathrm{T}}_{\mathrm{n}}({\mathrm{y}}_1,\dots, {\mathrm{y}}_{\mathrm{n}},\mathrm{y}))}^{\tau }$ where $\mathrm{M}{\models }\wedge_{1\le \mathrm{i}\le \mathrm{n}}\mathrm{I}({\mathrm{y}}_{\mathrm{i}})$ . Then

$$\begin{align*}\mathrm{M}\models \exists {\mathrm{u}}_1\dots {\mathrm{u}}_{\mathrm{n}}({\mathrm{u}}_1={\tau}_2({\mathrm{y}}_{\mathrm{n}},0)\ \&\ {\wedge}_{1\le \mathrm{i}<\mathrm{n}}{\mathrm{u}}_{\mathrm{i}+1}={\tau}_2({\mathrm{y}}_{\mathrm{n}-\mathrm{i}},{\mathrm{u}}_{\mathrm{i}})\ \&\ \mathrm{y}={\tau}_1({\mathrm{u}}_{\mathrm{n}})).\end{align*}$$

Suppose further that $\mathrm{M}\models \mathrm{x}\sqsubseteq \mathrm{y}$ where $\mathrm{M}\models \mathrm{I}(\mathrm{x})$ . Then $\mathrm{M}\models \mathrm{x}\sqsubseteq {\tau}_1({\mathrm{u}}_{\mathrm{n}})$ , whence, by $(\mathrm{T}{4}_1)$ , $\mathrm{M}\models \mathrm{x}={\tau}_1({\mathrm{u}}_{\mathrm{n}})\;\mathrm{v}\;\mathrm{x}\sqsubseteq {\mathrm{u}}_{\mathrm{n}}$ , and further, $\mathrm{M}\models \mathrm{x}=\mathrm{y}\kern0.24em \mathrm{v}\kern0.24em \mathrm{x}\sqsubseteq {\mathrm{u}}_{\mathrm{n}}$ . We now show, for $0\le \mathrm{j}<\mathrm{n}$ , that

(**) $$\begin{align}\mathrm{M}\models \mathrm{x}\sqsubseteq {\mathrm{u}}_{\mathrm{n}-\mathrm{j}}\to \mathrm{x}\sqsubseteq {\mathrm{y}}_{\mathrm{j}+1}\ \mathrm{v}\ \mathrm{x}\sqsubseteq {\mathrm{u}}_{\mathrm{n}-(\mathrm{j}+1)}.\end{align}$$

Assume that $\mathrm{M}\models \mathrm{x}\sqsubseteq {\mathrm{u}}_{\mathrm{n}-\mathrm{j}}$ . Then $\mathrm{M}\models \mathrm{x}\sqsubseteq {\tau}_2({\mathrm{y}}_{\mathrm{j}+1},{\mathrm{u}}_{\mathrm{n}-(\mathrm{j}+1)})$ , whence, by $(\mathrm{T}{4}_2)$ , $\mathrm{M}\models \mathrm{x}={\mathrm{u}}_{\mathrm{n}-\mathrm{j}}\;\mathrm{v}\;\mathrm{x}\sqsubseteq {\mathrm{y}}_{\mathrm{j}+1}\;\mathrm{v}\;\mathrm{x}\sqsubseteq {\mathrm{u}}_{\mathrm{n}-(\mathrm{j}+1)}$ . Suppose, for a reductio, that $\mathrm{M}\models \mathrm{x}={\mathrm{u}}_{\mathrm{n}-\mathrm{j}}$ . Then $\mathrm{M}\models \mathrm{x}={\tau}_2({\mathrm{y}}_{\mathrm{j}+1},{\mathrm{u}}_{\mathrm{n}-(\mathrm{j}+1)})$ . But from $\mathrm{M}\models \mathrm{I}(\mathrm{x})$ we have that $\mathrm{M}\models \exists \mathrm{z}\;\mathrm{x}={\tau}_1(\mathrm{z})$ , whence $\mathrm{M}\models {\tau}_2({\mathrm{y}}_{\mathrm{j}+1},{\mathrm{u}}_{\mathrm{n}-(\mathrm{j}+1)})={\tau}_1(\mathrm{z})$ , contradicting $(\mathrm{T}{2}_{1,2})$ . Hence $\mathrm{M}\models \mathrm{x}\ne {\mathrm{u}}_{\mathrm{n}-\mathrm{j}}$ . Therefore $\mathrm{M}\models \mathrm{x}\sqsubseteq {\mathrm{y}}_{\mathrm{j}+1}\;\mathrm{v}\;\mathrm{x}\sqsubseteq {\mathrm{u}}_{\mathrm{n}-(\mathrm{j}+1)}$ , as claimed. After $\mathrm{n}\hbox{--} 1$ applications of $(^{\ast \ast} )$ we obtain from $\mathrm{M}\models \mathrm{x}=\mathrm{y}\;\mathrm{v}\;\mathrm{x}\sqsubseteq {\mathrm{u}}_{\mathrm{n}}$ that $\mathrm{M}\models \mathrm{x}=\mathrm{y}\kern0.24em \mathrm{v}\kern0.24em {\vee}_{1\le \mathrm{i}\le \mathrm{n}-1}\mathrm{x}\sqsubseteq {\mathrm{y}}_{\mathrm{i}}\; \mathrm{v}\;\mathrm{x}\sqsubseteq {\mathrm{u}}_1$ . If $\mathrm{M}\models \mathrm{x}\sqsubseteq {\mathrm{u}}_1$ , then $\mathrm{M}\models \mathrm{x}\sqsubseteq {\tau}_2({\mathrm{y}}_{\mathrm{n}},0)$ , whence by $(\mathrm{T}{4}_2)$ , we have $\mathrm{M}\models \mathrm{x}={\tau}_2({\mathrm{y}}_{\mathrm{n}},0)\kern0.24em \mathrm{v}\;\;\mathrm{x}\sqsubseteq {\mathrm{y}}_{\mathrm{n}}\;\mathrm{v}\;\mathrm{x}\sqsubseteq 0$ . But $\mathrm{M}\models \mathrm{x}={\tau}_2({\mathrm{y}}_{\mathrm{n}},0)$ contradicts $\mathrm{M}\models \mathrm{I}(\mathrm{x})$ by $(\mathrm{T}{2}_{1,2})$ . And from $\mathrm{M}\models \mathrm{x}\sqsubseteq 0$ by $(\mathrm{T}6)$ and $(\mathrm{T}7)$ we obtain $\mathrm{M}\models \mathrm{x}=0$ . But then from $\mathrm{M}\models \mathrm{I}(\mathrm{x})$ we have $\mathrm{M}\models \exists \mathrm{z}\;{\tau}_1(\mathrm{z})=\mathrm{x}=0$ , contradicting $(\mathrm{T}{1}_1)$ . Hence both $\mathrm{M}\models \mathrm{x}={\tau}_2({\mathrm{y}}_{\mathrm{n}},0)$ and $\mathrm{M}\models \mathrm{x}\sqsubseteq 0$ are ruled out, and we obtain $\mathrm{M}\models \mathrm{x}=\mathrm{y}\kern0.24em \mathrm{v}\kern0.24em {\vee}_{1\le \mathrm{i}\le \mathrm{n}}\mathrm{x}\sqsubseteq {\mathrm{y}}_{\mathrm{i}}$ . But then

$$\begin{align*}\mathrm{M}\models \mathrm{x}\sqsubseteq \mathrm{y}\to \mathrm{x}=\mathrm{y}\kern0.24em \mathrm{v}\kern0.24em {\vee}_{1\le \mathrm{i}\le \mathrm{n}}\mathrm{x}\sqsubseteq {\mathrm{y}}_{\mathrm{i}}.\end{align*}$$

Conversely, assume that $\mathrm{M}\models \mathrm{x}=\mathrm{y}\kern0.24em \mathrm{v}\kern0.24em {\vee}_{1\le \mathrm{i}\le \mathrm{n}}\mathrm{x}\sqsubseteq {\mathrm{y}}_{\mathrm{i}}$ . Letting ${\mathrm{y}}_0=\mathrm{y}$ , we now show that the principal hypothesis $\mathrm{M}\models {({\mathrm{T}}_{\mathrm{n}}({\mathrm{y}}_1,\dots, {\mathrm{y}}_{\mathrm{n}},\mathrm{y}))}^{\tau }$ implies that

(†) $$\begin{align}\mathrm{M}\models\wedge_{1\le \mathrm{i}\le \mathrm{n}}{\mathrm{y}}_{\mathrm{i}}\sqsubseteq \mathrm{y}.\end{align}$$

By $(\mathrm{T}5)$ we have that $\mathrm{M}\models \mathrm{y}\sqsubseteq \mathrm{y}$ . Hence $\mathrm{M}\models {\mathrm{y}}_0\sqsubseteq \mathrm{y}$ . By $(\mathrm{T}{4}_1)$ we have that $\mathrm{M}\models {\mathrm{u}}_{\mathrm{n}}\sqsubseteq {\tau}_1({\mathrm{u}}_{\mathrm{n}})=\mathrm{y}$ . Next, we show, for $0\le \mathrm{j}<\mathrm{n}$ , that

(***) $$\begin{align}\mathrm{M}\models {\mathrm{y}}_{\mathrm{j}}\sqsubseteq \mathrm{y}\ \&\ {\mathrm{u}}_{\mathrm{n}-\mathrm{j}}\sqsubseteq \mathrm{y}\to {\mathrm{y}}_{\mathrm{j}+1}\sqsubseteq \mathrm{y}\ \&\ {\mathrm{u}}_{\mathrm{n}-(\mathrm{j}+1)}\sqsubseteq \mathrm{y}.\end{align}$$

Assume that $\mathrm{M}\models {\mathrm{y}}_{\mathrm{j}}\sqsubseteq \mathrm{y}\ \&\ {\mathrm{u}}_{\mathrm{n}-\mathrm{j}}\sqsubseteq \mathrm{y}$ . Then, by $(\mathrm{T}{4}_2)$ , we have that

$\mathrm{M}\models {\mathrm{y}}_{\mathrm{j}+1}\sqsubseteq {\tau}_2({\mathrm{y}}_{\mathrm{j}+1},{\mathrm{u}}_{\mathrm{n}-(\mathrm{j}+1)})={\mathrm{u}}_{\mathrm{n}-\mathrm{j}}\sqsubseteq \mathrm{y}$ and $\mathrm{M}\models {\mathrm{u}}_{\mathrm{n}-(\mathrm{j}+1)}\sqsubseteq {\tau}_2({\mathrm{y}}_{\mathrm{j}},$ ${\mathrm{u}}_{\mathrm{n}-(\mathrm{j}+1)})={\mathrm{u}}_{\mathrm{n}-\mathrm{j}}\sqsubseteq \mathrm{y}$ .

This proves $(^{\ast \ast \ast} )$ .

After $\mathrm{n}$ applications of $(^{\ast \ast \ast} )$ we then obtain from $\mathrm{M}\models {\mathrm{y}}_0\sqsubseteq \mathrm{y}\ \&\ {\mathrm{u}}_{\mathrm{n}}\sqsubseteq \mathrm{y}$ that

$$\begin{align*}\mathrm{M}\models \wedge_{1\le \mathrm{i}<\mathrm{n}}{\mathrm{y}}_{\mathrm{i}}\sqsubseteq \mathrm{y}\ \&\ {\mathrm{u}}_1\sqsubseteq \mathrm{y}.\end{align*}$$

But then, by $(\mathrm{T}{4}_2)$ , $\mathrm{M}\models {\mathrm{y}}_{\mathrm{n}}\sqsubseteq {\tau}_2({\mathrm{y}}_{\mathrm{n}},0)={\mathrm{u}}_1\sqsubseteq \mathrm{y}$ . Hence by $(\mathrm{T}8)$ it follows that

$$\begin{align*}\mathrm{M} \models \wedge_{1\le \mathrm{i}\le \mathrm{n}}{\mathrm{y}}_{\mathrm{i}}\sqsubseteq \mathrm{y}.\end{align*}$$

This establishes $(\dagger )$ . But then from hypothesis $\mathrm{M}\models \mathrm{x}=\mathrm{y}\kern0.24em \mathrm{v}\vee {}_{1\le \mathrm{i}\le \mathrm{n}}\mathrm{x}\sqsubseteq {\mathrm{y}}_{\mathrm{i}}$ and ${\mathrm{y}}_0=\mathrm{y}$ , we obtain, by $(\mathrm{T}8)$ , that $\mathrm{M}\models \mathrm{x}\sqsubseteq \mathrm{y}$ . Therefore also

$$\begin{align*}\mathrm{M}\models \mathrm{x}=\mathrm{y}\kern0.24em \mathrm{v}\kern0.24em {\vee}_{1\le \mathrm{i}\le \mathrm{n}}\mathrm{x}\sqsubseteq {\mathrm{y}}_{\mathrm{i}}\to \mathrm{x}\sqsubseteq \mathrm{y},\end{align*}$$

as required.

(e) Assume $\mathrm{M}\models \wedge_{1\le \mathrm{i}\le \mathrm{n}}\mathrm{I}({\mathrm{x}}_{\mathrm{i}})$ . We have that $\mathrm{M}\models \exists {\mathrm{u}}_1\kern0.24em {\mathrm{u}}_1={\tau}_2({\mathrm{x}}_{\mathrm{n}},0)$ .

Let ${\mathrm{u}}_0=0$ . We now argue that, for $0\le \mathrm{j} - 1<\mathrm{n}$ ,

(††) $$\begin{align}\mathrm{M}\models {\mathrm{u}}_{\mathrm{j}+1}={\tau}_2({\mathrm{x}}_{\mathrm{n}-\mathrm{j}},{\mathrm{u}}_{\mathrm{j}})\to \exists {\mathrm{u}}_{\mathrm{j}+2}{\mathrm{u}}_{\mathrm{j}+2}={\tau}_2({\mathrm{x}}_{\mathrm{n}-(\mathrm{j}+1)},{\mathrm{u}}_{\mathrm{j}+1}).\end{align}$$

Assume that $\mathrm{M}\models {\mathrm{u}}_{\mathrm{j}+1}={\tau}_2({\mathrm{x}}_{\mathrm{n}-\mathrm{j}},{\mathrm{u}}_{\mathrm{j}})$ . Then, trivially in ${\mathrm{T}}_{\le 2}$ ,

$$\begin{align*}\mathrm{M}\models \exists {\mathrm{u}}_{\mathrm{j}+2}{\mathrm{u}}_{\mathrm{j}+2}={\tau}_2({\mathrm{x}}_{\mathrm{n}-(\mathrm{j}+1)},{\mathrm{u}}_{\mathrm{j}+1}).\end{align*}$$

Hence $(\dagger \dagger )$ holds. After $\mathrm{n}\hbox{--} 1$ applications of $(\dagger \dagger )$ we obtain from $\mathrm{M}\models \exists {\mathrm{u}}_1{\mathrm{u}}_1={\tau}_2({\mathrm{x}}_{\mathrm{n}},0)$ that

$$\begin{align*}\mathrm{M}\models \exists {\mathrm{u}}_1\dots {\mathrm{u}}_{\mathrm{n}}({\mathrm{u}}_1={\tau}_2({\mathrm{x}}_{\mathrm{n}},0)\ \&\ {\wedge}_{1\le \mathrm{i}<\mathrm{n}}{\mathrm{u}}_{\mathrm{i}+1}={\tau}_2({\mathrm{x}}_{\mathrm{n}-\mathrm{i}},{\mathrm{u}}_{\mathrm{i}})).\end{align*}$$

Now, we also have that $\mathrm{M}\models \exists \mathrm{x}\;\mathrm{x}={\tau}_1({\mathrm{u}}_{\mathrm{n}})$ . From the definition of $\mathrm{I}$ we have that $\mathrm{M}\models \exists \mathrm{x}\;(\mathrm{x}={\tau}_1({\mathrm{u}}_{\mathrm{n}})\ \&\ \mathrm{I}(\mathrm{x}))$ . Hence we have proved that $\mathrm{M}\models \exists \mathrm{x}\;(\mathrm{I}(\mathrm{x})\ \&\ {({\mathrm{T}}_{\mathrm{n}}({\mathrm{x}}_1,\dots, {\mathrm{x}}_{\mathrm{n}},\mathrm{x}))}^{\tau})$ , as required.

(f) Assume $\mathrm{M}\models {({\mathrm{T}}_{\mathrm{n}}({\mathrm{x}}_1,\dots, {\mathrm{x}}_{\mathrm{n}},\mathrm{x}))}^{\tau}\ \&\ {({\mathrm{T}}_{\mathrm{n}}({\mathrm{x}}_1,\dots, {\mathrm{x}}_{\mathrm{n}},\mathrm{y}))}^{\tau }$ where $\mathrm{M}\models \wedge_{1\le \mathrm{i}\le \mathrm{n}}\mathrm{I}({\mathrm{x}}_{\mathrm{i}})\ \&\ \mathrm{I}(\mathrm{x})\ \&\ \mathrm{I}(\mathrm{y})$ . Then $\mathrm{M}\models \exists {\mathrm{u}}_1\dots {\mathrm{u}}_{\mathrm{n}}({\mathrm{u}}_1={\tau}_2({\mathrm{x}}_{\mathrm{n}},0)\ \& $ ${\wedge}_{1\le \mathrm{i}<\mathrm{n}}\ {\mathrm{u}}_{\mathrm{i}+1}={\tau}_2({\mathrm{x}}_{\mathrm{n}-\mathrm{i}},{\mathrm{u}}_{\mathrm{i}})\ \&\ \mathrm{x}={\tau}_1({\mathrm{u}}_{\mathrm{n}}))$ and $\mathrm{M}\models \exists {\mathrm{v}}_1\dots {\mathrm{v}}_{\mathrm{n}}({\mathrm{v}}_1={\tau}_2({\mathrm{x}}_{\mathrm{n}},0)\ \& $ ${\wedge}_{1\le \mathrm{i}<\mathrm{n}}\ {\mathrm{v}}_{\mathrm{i}+1}={\tau}_2({\mathrm{x}}_{\mathrm{n}-\mathrm{i}},{\mathrm{v}}_{\mathrm{i}})\ \&\ \mathrm{y}={\tau}_1({\mathrm{v}}_{\mathrm{n}}))$ . Hence $\mathrm{M}\models {\mathrm{u}}_1={\mathrm{v}}_1$ . Now, for $1\le \mathrm{j}<\mathrm{n}$ , we have that

(†††) $$\begin{align}\mathrm{M}\models {\mathrm{u}}_{\mathrm{j}}={\mathrm{v}}_{\mathrm{j}}\to {\mathrm{u}}_{\mathrm{j}+1}={\mathrm{v}}_{\mathrm{j}+1}.\end{align}$$

After $\mathrm{n}\hbox{--} 1$ applications of $(\dagger \dagger \dagger )$ we obtain from $\mathrm{M}\models {\mathrm{u}}_1={\mathrm{v}}_1$ that $\mathrm{M}\models {\mathrm{u}}_{\mathrm{n}}={\mathrm{v}}_{\mathrm{n}}$ . But then $\mathrm{M}\models \mathrm{x}={\tau}_1({\mathrm{u}}_{\mathrm{n}})={\tau}_1({\mathrm{v}}_{\mathrm{n}})=\mathrm{y}$ . So $\mathrm{M}\models \mathrm{x}=\mathrm{y}$ , as required.

Proof For (a), see [Reference Damnjanovic2, 4.17(b)]; in [Reference Damnjanovic3] the formula $\mathrm{I}^\ast (\mathrm{x})$ is selected so as to ensure this property. Part (b) is proved as $(\mathrm{I}1)$ in [Reference Damnjanovic3, Section 6.1(c)]. Parts (c) and (d) are straightforwardly obtained from the definition of $\mathrm{I}^\ast$ in [Reference Damnjanovic3] (cf. 2.3(a) and (c)). For (e), assume $\mathrm{M}\models \mathrm{I}^\ast (\mathrm{y})\ \&\ \mathrm{I}^\ast (\mathrm{z})$ . Then $\mathrm{M}\models \mathrm{J}^\ast (\mathrm{y})\ \&\ \mathrm{J}^\ast (\mathrm{z})$ , where $\mathrm{J}^\ast$ is the string form selected in [Reference Damnjanovic3, Section 6] that is closed under ${}^\ast$ and downward closed under ${\subseteq}_{\mathrm{p}}$ . Hence $\mathrm{M}\models \mathrm{J}^\ast (\mathrm{bbayz})$ . The claim follows from the fact that $\mathrm{J}^\ast$ was also selected to have the property described in (3.10) of [Reference Damnjanovic2]. The same proof for (f) and (g) taking into account (3.6) and (3.12) of [Reference Damnjanovic2].

Proof For (a), assume $\mathrm{M}\models {\mathrm{I}}_{\le 2}(\mathrm{x})\ \&\ \mathrm{z}=\mathrm{bax}$ . Then $\mathrm{M}\models \mathrm{I}^\ast (\mathrm{x})$ . We have that $\mathrm{M}\models \mathrm{I}^\ast (\mathrm{a})$ , by definition of $\mathrm{I}^\ast$ . By 5.1(b) we then have that $\mathrm{M}\models \mathrm{I}^\ast (\mathrm{z})$ .

Assume now that $\mathrm{M}\models {\mathrm{Tally}}_{\mathrm{b}}(\mathrm{t})\ \&\ \mathrm{t}{\subseteq}_{\mathrm{p}}\mathrm{z}$ . Then $\mathrm{M}\models \mathrm{t}{\subseteq}_{\mathrm{p}}\mathrm{bax}$ . But then, by 5.1(a) we have that $\mathrm{M}\models \mathrm{t}{\subseteq}_{\mathrm{p}}\mathrm{b}\;\mathrm{v}\;\mathrm{t}{\subseteq}_{\mathrm{p}}\mathrm{x}$ . Now, from $\mathrm{M}\models \mathrm{t}{\subseteq}_{\mathrm{p}}\mathrm{b}$ we have $\mathrm{M}\models \mathrm{t}=\mathrm{b}$ , and from $\mathrm{M}\models \mathrm{t}{\subseteq}_{\mathrm{p}}\mathrm{x}$ and hypothesis $\mathrm{M}\models {\mathrm{I}}_{\le 2}(\mathrm{x})$ we have $\mathrm{M}\models \mathrm{t}=\mathrm{b}\;\mathrm{v}\;\mathrm{t}=\mathrm{b}\mathrm{b}$ . Hence $\mathrm{M}\models \mathrm{t}=\mathrm{b}\;\mathrm{v}\;\mathrm{t}=\mathrm{b}\mathrm{b}$ , as required.

For (b), assume $\mathrm{M}\models {\mathrm{I}}_{\le 2}(\mathrm{x})\ \&\ {\mathrm{I}}_{\le 2}(\mathrm{y})\ \&\ \mathrm{z}=\mathrm{bbaxy}$ . Then $\mathrm{M}\models {\mathrm{I}}_{\le 2}(\mathrm{bax})$ by (a), so $\mathrm{M}\models \mathrm{I}^\ast (\mathrm{bax})$ . From hypothesis $\mathrm{M}\models {\mathrm{I}}_{\le 2}(\mathrm{y})$ we have $\mathrm{M}\models \mathrm{I}^\ast (\mathrm{y})$ . But then $\mathrm{M}\models \mathrm{I}^\ast (\mathrm{z})$ by 5.1(b).

Assume now that $\mathrm{M}\models {\mathrm{Tally}}_{\mathrm{b}}(\mathrm{t})\ \&\ \mathrm{t}{\subseteq}_{\mathrm{p}}\mathrm{z}$ . Then $\mathrm{M}\models \mathrm{t}{\subseteq}_{\mathrm{p}}\mathrm{bbaxy}$ . By 5.1(a), we have that $\mathrm{M}\models \mathrm{t}{\subseteq}_{\mathrm{p}}\mathrm{bb}\;\mathrm{v}\;\mathrm{t}{\subseteq}_{\mathrm{p}}\mathrm{xy}$ . If $\mathrm{M}\models \mathrm{t}{\subseteq}_{\mathrm{p}}\mathrm{bb}$ , then $\mathrm{M}\models \mathrm{t}=\mathrm{b}\;\mathrm{v}\;\mathrm{t}=\mathrm{b}\mathrm{b}$ . Suppose now that $\mathrm{M}\models \mathrm{t}{\subseteq}_{\mathrm{p}}\mathrm{xy}$ . From $\mathrm{M}\models \mathrm{I}^\ast (\mathrm{x})$ we have that $\mathrm{M}\models \mathrm{x}=\mathrm{a}\;\mathrm{v}\;\mathrm{a}\mathrm{aEx}$ . If $\mathrm{M}\models \mathrm{x}=\mathrm{a}$ , then from $\mathrm{M}\models \mathrm{t}{\subseteq}_{\mathrm{p}}\mathrm{ay}$ we have $\mathrm{M}\models \mathrm{t}{\subseteq}_{\mathrm{p}}\mathrm{y}$ .

If $\mathrm{M}\models \mathrm{aaEx}$ , then $\mathrm{M}\models \exists {\mathrm{x}}_1\mathrm{x}={\mathrm{x}}_1\mathrm{aa}$ , so $\mathrm{M}\models \mathrm{t}{\subseteq}_{\mathrm{p}}({\mathrm{x}}_1\mathrm{aa})\mathrm{y}$ , whence $\mathrm{M}\models \mathrm{t}{\subseteq}_{\mathrm{p}}{\mathrm{x}}_1\;\mathrm{v}\kern0.24em \mathrm{t}{\subseteq}_{\mathrm{p}}\mathrm{y}$ . So from $\mathrm{M}\models \mathrm{t}{\subseteq}_{\mathrm{p}}\mathrm{xy}$ we have $\mathrm{M}\models \mathrm{t}{\subseteq}_{\mathrm{p}}\mathrm{x}\;\mathrm{v}\;\mathrm{t}{\subseteq}_{\mathrm{p}}\mathrm{y}$ , and thus from hypothesis $\mathrm{M}\models {\mathrm{I}}_{\le 2}(\mathrm{x})\ \&\ {\mathrm{I}}_{\le 2}(\mathrm{y})$ we obtain $\mathrm{M}\models \mathrm{t}=\mathrm{b}\;\mathrm{v}\;\mathrm{t}=\mathrm{b}\mathrm{b}$ , as required.

Proof (a) We show that

$\mathrm{QT}^+ {\kern-1.5pt}\vdash{\kern-1.5pt} \mathrm{I}_{\leq 2}(\mathrm{x}) \, \&\, \mathrm{I}_{\leq 2}(\mathrm{y}) {\kern-1.5pt}\rightarrow{\kern-1.5pt} (\mathrm{x}{\kern-1.5pt} ={\kern-1.5pt} \mathrm{bay}\, \mathrm{v}\, \mathrm{ax} \subseteq_\mathrm{p}\kern-3pt{\mathrm{bay}} \leftrightarrow \mathrm{x} = \mathrm{bay}\, \mathrm{v}\, \mathrm{x} {\kern-1.5pt}={\kern-1.5pt} \mathrm{y}\, \mathrm{v}\, \mathrm{ax}{\subseteq}_{\mathrm{p}}{\mathrm{y}}{\kern-1pt}). $ Assume $\mathrm{M}\models {\mathrm{I}}_{\le 2}(\mathrm{x})\ \&\ {\mathrm{I}}_{\le 2}(\mathrm{y})$ . Suppose that $\mathrm{M}\models \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{bay}$ . We can rule out $\mathrm{M}\models \mathrm{ax}=\mathrm{bay}\;\mathrm{v}\kern0.24em \mathrm{axBbay}$ immediately. Then we have $\mathrm{M}\models \mathrm{axEbay}\kern0.24em \mathrm{v}\ \exists {\mathrm{x}}_1,{\mathrm{x}}_2\;\mathrm{bay}={\mathrm{x}}_1(\mathrm{ax}){\mathrm{x}}_2$ , that is,

$$\begin{align*}\mathrm{M}\models \exists {\mathrm{x}}_1\mathrm{bay}={\mathrm{x}}_1(\mathrm{ax})\kern0.24em \mathrm{v}\ \exists {\mathrm{x}}_1,{\mathrm{x}}_2\;\mathrm{bay}={\mathrm{x}}_1(\mathrm{ax}){\mathrm{x}}_2.\end{align*}$$

Hence $\mathrm{M}\models \exists {\mathrm{x}}_1{\mathrm{x}}_1\mathrm{Bbay}$ . Now, we have that

$$\begin{align*}\mathrm{QT}^{+}\vdash \mathrm{zBbaw}\rightarrow \mathrm{z} = \mathrm{b\ v\ z} = \mathrm{ba\ v}\ \mathrm{baBz}\end{align*}$$

So $\mathrm{M}\models {\mathrm{x}}_1=\mathrm{b}\kern0.24em \mathrm{v}\kern0.24em {\mathrm{x}}_1=\mathrm{b}\mathrm{a}\;\mathrm{v}\;{\mathrm{baBx}}_1$ . We distinguish the cases:

$$\begin{align*}\hspace{-136pt}(1)\hspace{136pt}\mathrm{M}\models {\mathrm{x}}_1=\mathrm{b}.\end{align*}$$

Then $\mathrm{M}\models \mathrm{bay}=\mathrm{b}\mathrm{ax}\kern0.24em \mathrm{v}\kern0.24em \mathrm{bay}=\mathrm{b}(\mathrm{ax}){\mathrm{x}}_2$ , whence $\mathrm{M}\models \mathrm{y}=\mathrm{x}\kern0.24em \mathrm{v}\kern0.24em \mathrm{y}={\mathrm{xx}}_2$ , so $\mathrm{M}\models \mathrm{y}=\mathrm{x}\;\mathrm{v}\;\mathrm{x}\mathrm{By}$ . But $\mathrm{M}\models \mathrm{xBy}$ is ruled out by $\mathrm{M}\models \mathrm{I}^\ast (\mathrm{x})\ \&\ \mathrm{I}^\ast (\mathrm{y})$ and 5.1(c). Hence $\mathrm{M}\models \mathrm{x}=\mathrm{y}$ .

$$\begin{align*}\hspace{-135pt}(2)\hspace{135pt}\mathrm{M}\models {\mathrm{x}}_1=\mathrm{ba}.\end{align*}$$

Then $\mathrm{M}\models \mathrm{bay}=\mathrm{ba}(\mathrm{ax})\kern0.24em \mathrm{v}\;\;\mathrm{ba}\mathrm{y}=\mathrm{ba}(\mathrm{ax}){\mathrm{x}}_2$ , whence $\mathrm{M}\models \mathrm{y}=\mathrm{ax}\kern0.24em \mathrm{v}\kern0.24em \mathrm{y}={\mathrm{axx}}_2$ . But $\mathrm{M}\models \mathrm{aBy}$ contradicts $\mathrm{M}\models \mathrm{I}^\ast (\mathrm{y})$ . So this case is ruled out.

$$\begin{align*}\hspace{-138.5pt}(3)\hspace{138.5pt}\mathrm{M}\models {\mathrm{baBx}}_1.\end{align*}$$

Then $\mathrm{M}\models \exists {\mathrm{x}}_3{\mathrm{x}}_1={\mathrm{bax}}_3$ , so $\mathrm{M}\models \mathrm{bay}={\mathrm{bax}}_3(\mathrm{ax})\kern0.24em \mathrm{v}\;\;\mathrm{bay}={\mathrm{bax}}_3(\mathrm{ax}){\mathrm{x}}_2$ , whence $\mathrm{M}\models \mathrm{y}={\mathrm{x}}_3(\mathrm{ax})\;\mathrm{v}\;\mathrm{y}={\mathrm{x}}_3(\mathrm{ax}){\mathrm{x}}_2$ . But then $\mathrm{M}\models \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{y}$ .

Therefore, $\mathrm{M}\models \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{bay}\to \mathrm{x}=\mathrm{y}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{y}$ . Conversely, suppose $\mathrm{M}\models \mathrm{x}=\mathrm{y}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{y}$ . Then $\mathrm{M}\models \mathrm{ax}=\mathrm{ay}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{y}$ , whence $\mathrm{M}\models \mathrm{axEbay}\kern0.24em \mathrm{v} \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{bay}$ , which yields $\mathrm{M}\models \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{bay}$ . Therefore we also have $\mathrm{M}\models \mathrm{x}=\mathrm{y}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{y}\to \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{bay}$ . But then $\mathrm{M}\models \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{bay}\leftrightarrow \mathrm{x}=\mathrm{y}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{y}$ , and a fortiori, $\mathrm{M}\models \mathrm{x}=\mathrm{bay}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{bay}\leftrightarrow \mathrm{x}=\mathrm{bay}\kern0.24em \mathrm{v}\kern0.24em \mathrm{x}=\mathrm{y}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{y}$ , as needed.

(b) We show that

$$\begin{align*}&\mathrm{QT}^{+}\vdash \mathrm{I}_{\leq2}(\mathrm{x}) \ \&\ \mathrm{I}_{\leq2} (\mathrm{y})\ \&\ \mathrm{I}_{\leq2} (\mathrm{z}) \rightarrow\\&\rightarrow (\mathrm{x} = \mathrm{bbayz}\ \mathrm{v}\ \mathrm{ax} \subseteq_{\mathrm{p}} \mathrm{bbayz} \leftrightarrow \mathrm{x} =\mathrm{bbayz}\ \mathrm{v}\\&\qquad \mathrm{v}\ (\mathrm{x} = \mathrm{y\ v\ ax} \subseteq_{\mathrm{p}} \mathrm{y}) \,\mathrm{v}\ (\mathrm{x} = \mathrm{z\ v\ ax}\subseteq_{\mathrm{p}} \mathrm{z})).\end{align*}$$

Assume $\mathrm{M}\models {\mathrm{I}}_{\le 2}(\mathrm{x})\ \&\ {\mathrm{I}}_{\le 2}(\mathrm{y})\ \&\ {\mathrm{I}}_{\le 2}(\mathrm{z})$ . Suppose that $\mathrm{M}\models \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{bbayz}$ . Again we can rule out $\mathrm{M}\models \mathrm{ax}=\mathrm{bbayz}\kern0.24em \mathrm{v}\kern0.24em \mathrm{axB}(\mathrm{bbayz})$ . So we are left with $\mathrm{M}\models \mathrm{axE}(\mathrm{bbayz})\kern0.24em \mathrm{v}\ \exists {\mathrm{x}}_1,{\mathrm{x}}_2\;\mathrm{bbayz}={\mathrm{x}}_1(\mathrm{ax}){\mathrm{x}}_2$ .

$$\begin{align*}\hspace{-125pt}(1)\hspace{125pt}\mathrm{M}\models \mathrm{axE}(\mathrm{bbayz}).\end{align*}$$

Then by 5.1(e), $\mathrm{M}\models \mathrm{axEz}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}=\mathrm{z}\kern0.24em \mathrm{v}\kern0.24em \mathrm{zEax}$ . If $\mathrm{M}\models \mathrm{axEz}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}=\mathrm{z}$ , then $\mathrm{M}\models \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{z}$ , and we are done. So we may assume that $\mathrm{M}\models \mathrm{zEax}$ . Then $\mathrm{M}\models \exists {\mathrm{x}}_1\mathrm{ax}={\mathrm{x}}_1\mathrm{z}$ , whence $\mathrm{M}\models {\mathrm{x}}_1=\mathrm{a}\kern0.24em \mathrm{v}\kern0.24em {\mathrm{aBx}}_1$ . If $\mathrm{M}\models {\mathrm{x}}_1=\mathrm{a}$ , then $\mathrm{M}\models \mathrm{ax}=\mathrm{az}$ , and we obtain $\mathrm{M}\models \mathrm{x}=\mathrm{z}$ , as needed. If $\mathrm{M}\models {\mathrm{aBx}}_1$ , then $\mathrm{M}\models \exists {\mathrm{x}}_2\ \mathrm{ax}={\mathrm{x}}_1\mathrm{z}=({\mathrm{ax}}_2)\mathrm{z}$ , whence $\mathrm{M}\models \mathrm{x}={\mathrm{x}}_2\mathrm{z}$ , and $\mathrm{M}\models {\mathrm{x}}_2\mathrm{Bx}$ .

From hypothesis $\mathrm{M}\models \mathrm{axE}(\mathrm{bbayz})$ , we have $\mathrm{M}\models \exists {\mathrm{z}}_1\;\mathrm{bbayz}={\mathrm{z}}_1\mathrm{a}\mathrm{x}={\mathrm{z}}_1\mathrm{a}({\mathrm{x}}_2\mathrm{z})$ .

By 5.1(f), we obtain $\mathrm{M}\models \mathrm{bbay}={\mathrm{z}}_1{\mathrm{ax}}_2$ , whence $\mathrm{M}\models {\mathrm{z}}_1=\mathrm{b}\kern0.24em \mathrm{v}\kern0.24em {\mathrm{bBz}}_1$ . It follows that $\mathrm{M}\models \mathrm{bbay}={\mathrm{bax}}_2\;\mathrm{v}\kern0.24em \exists {\mathrm{z}}_2\;\mathrm{bbay}=({\mathrm{bz}}_2){\mathrm{ax}}_2$ , hence also $\mathrm{M}\models \mathrm{bay}={\mathrm{ax}}_2\;\mathrm{v}\kern0.24em \mathrm{bay}={\mathrm{z}}_2{\mathrm{ax}}_2$ . But $\mathrm{M}\models \mathrm{bay}={\mathrm{ax}}_2$ is ruled out. Hence $\mathrm{M}\models \mathrm{bay}={\mathrm{z}}_2{\mathrm{ax}}_2$ , and so $\mathrm{M}\models {\mathrm{x}}_2\mathrm{Ebay}$ . Now by 5.1(b), we have that $\mathrm{M}\models \mathrm{I}^\ast (\mathrm{bay})$ since $\mathrm{M}\models \mathrm{I}^\ast (\mathrm{a})\ \&\ \mathrm{I}^\ast (\mathrm{y})$ . But then $\mathrm{M}\models {\mathrm{x}}_2\mathrm{Bx}\ \&\ {\mathrm{x}}_2\mathrm{Ebay}$ contradicts $\mathrm{M}\models \mathrm{I}^\ast (\mathrm{x})\ \&\ \mathrm{I}^\ast (\mathrm{y})$ by 5.1(d). Hence subcase $\mathrm{M}\models {\mathrm{aBx}}_1$ is ruled out.

$$\begin{align*}\hspace{-95pt}(2)\hspace{95pt}\mathrm{M}\models \exists {\mathrm{x}}_1,{\mathrm{x}}_2\;\mathrm{bbayz}={\mathrm{x}}_1(\mathrm{ax}){\mathrm{x}}_2.\end{align*}$$

Then by 5.1(e), $\mathrm{M}\models {\mathrm{x}}_2\mathrm{Ez}\kern0.24em \mathrm{v}\kern0.24em {\mathrm{x}}_2=\mathrm{z}\;\mathrm{v}\;{\mathrm{zEx}}_2$ .

$$\begin{align*}\hspace{-120pt}(2\mathrm{a})\hspace{120pt}\mathrm{M}\models {\mathrm{zEx}}_2\mathrm{v}\;{\mathrm{x}}_2=\mathrm{z}.\end{align*}$$

Then $\mathrm{M}\models \exists {\mathrm{x}}_4{\mathrm{x}}_2={\mathrm{x}}_4\mathrm{z}\kern0.24em \mathrm{v}\kern0.24em {\mathrm{x}}_2=\mathrm{z}$ , so $\mathrm{M}\models \mathrm{bbayz}={\mathrm{x}}_1\mathrm{ax}({\mathrm{x}}_4\mathrm{z})\kern0.24em \mathrm{v}\;\;\mathrm{bbayz}={\mathrm{x}}_1(\mathrm{ax})\mathrm{z}$ .

By 5.1(f), we have $\mathrm{M}\models \mathrm{bbay}={\mathrm{x}}_1{\mathrm{axx}}_4\;\mathrm{v}\kern0.24em \mathrm{bbay}={\mathrm{x}}_1(\mathrm{ax})$ , whence $\mathrm{M}\models {\mathrm{x}}_1=\mathrm{b}\;\mathrm{v}\;{\mathrm{bBx}}_1$ . It follows that

$$\begin{align*}\mathrm{M}\models \mathrm{bbay}={\mathrm{baxx}}_4\;\mathrm{v}\ \exists {\mathrm{x}}_3\mathrm{bbay}=({\mathrm{bx}}_3){\mathrm{axx}}_4\;\mathrm{v}\kern0.24em \mathrm{bbay}=\mathrm{b}(\mathrm{ax})\;\;\mathrm{v}\ \exists {\mathrm{x}}_3\mathrm{bbay}=({\mathrm{bx}}_3)\mathrm{ax},\end{align*}$$

Hence $\mathrm{M}\models \mathrm{bay}={\mathrm{axx}}_4\;\mathrm{v}\kern0.24em \mathrm{bay}={\mathrm{x}}_3{\mathrm{axx}}_4\;\mathrm{v}\kern0.24em \mathrm{bay}=\mathrm{ax}\kern0.24em \mathrm{v}\kern0.24em \mathrm{bay}={\mathrm{x}}_3\mathrm{ax}$ , whence $\mathrm{M}\models \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{bay}$ . Then $\mathrm{M}\models \mathrm{x}=\mathrm{y}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{y}$ follows as in 5.3(a).

$$\begin{align*}\hspace{-139pt}(2\mathrm{b})\hspace{139pt}\mathrm{M}\models {\mathrm{x}}_2\mathrm{Ez}.\end{align*}$$

Then $\mathrm{M}\models \exists {\mathrm{x}}_3\ \mathrm{z}={\mathrm{x}}_3{\mathrm{x}}_2$ ; thus $\mathrm{M}\models {\mathrm{x}}_3\mathrm{Bz}$ . Also, $\mathrm{M}\models \mathrm{bbay}({\mathrm{x}}_3{\mathrm{x}}_2)={\mathrm{x}}_1(\mathrm{ax}){\mathrm{x}}_2$ . Hence, by 5.1(f), $\mathrm{M}\models {\mathrm{bbayx}}_3={\mathrm{x}}_1(\mathrm{ax})$ . Then, by 5.1(e), $\mathrm{M}\models {\mathrm{x}}_3\mathrm{Ex}\kern0.24em \mathrm{v}\kern0.24em {\mathrm{x}}_3=\mathrm{x}\kern0.24em \mathrm{v}\kern0.24em {\mathrm{x}\mathrm{Ex}}_3$ .

$$\begin{align*}\hspace{-116pt}(2\mathrm{bi})\hspace{116pt}\mathrm{M}\models {\mathrm{x}}_3\mathrm{Ex}\kern0.24em \mathrm{v}\kern0.24em {\mathrm{x}}_3=\mathrm{x}.\end{align*}$$

Then either way from $\mathrm{M}\models {\mathrm{x}}_3\mathrm{Bz}$ and $\mathrm{M}\models \mathrm{I}^\ast (\mathrm{x})\ \&\ \mathrm{I}^\ast (\mathrm{z})$ we obtain a contradiction by 5.1(c) and 5.1(d). Hence subcase (2bi) is ruled out.

$$\begin{align*}\hspace{-134pt}(2\mathrm{bii})\hspace{134pt}\mathrm{M}\models {\mathrm{xEx}}_3.\end{align*}$$

Then $\mathrm{M}\models \exists {\mathrm{x}}_5\ {\mathrm{x}}_3={\mathrm{x}}_5\mathrm{x}$ , whence $\mathrm{M}\models {\mathrm{bbayx}}_5\mathrm{x}={\mathrm{x}}_1\mathrm{ax}$ . By 5.1(f), we have $\mathrm{M}\models {\mathrm{bbayx}}_5={\mathrm{x}}_1\mathrm{a}$ . Hence $\mathrm{M}\models {\mathrm{x}}_5=\mathrm{a}\;\mathrm{v}\;{\mathrm{aEx}}_5$ . Then from $\mathrm{M}\models {\mathrm{x}}_3={\mathrm{x}}_5\mathrm{x}$ , we obtain $\mathrm{M}\models {\mathrm{x}}_3=\mathrm{ax}\;\mathrm{v}\kern0.24em {\mathrm{axEx}}_3$ . From $\mathrm{M}\models {\mathrm{x}}_3\mathrm{Bz}$ , we obtain $\mathrm{M}\models \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{z}$ , as needed.

This proves that $\mathrm{M}\models \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{bbayz}\to (\mathrm{x}=\mathrm{y}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{y})\;\mathrm{v}\;(\mathrm{x}=\mathrm{z}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{z})$ .

Conversely, suppose that $\mathrm{M}\models \mathrm{x}=\mathrm{y}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{y}$ . If $\mathrm{M}\models \mathrm{x}=\mathrm{y}$ , then $\mathrm{M}\models \mathrm{ax}=\mathrm{ay}$ , whence $\mathrm{M}\models \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{bb}(\mathrm{ay})\mathrm{z}$ . So $\mathrm{M}\models \mathrm{x}=\mathrm{y}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{y}\to \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{bbayz}$ . Likewise $\mathrm{M}\models \mathrm{x}=\mathrm{z}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{z}\to \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{bbayz}$ . Therefore we also have

$$\begin{align*}\mathrm{M}\models (\mathrm{x}=\mathrm{y}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{y})\;\mathrm{v}\;(\mathrm{x}=\mathrm{z}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{z})\to \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{bbayz}.\end{align*}$$

But then $\mathrm{M}\models \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{bbayz}\leftrightarrow (\mathrm{x}=\mathrm{y}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{y})\;\mathrm{v}\;(\mathrm{x}=\mathrm{z}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{z})$ , and finally $\mathrm{M}\models \mathrm{x}=\mathrm{bbayz}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{bbayz}\leftrightarrow \mathrm{x}=\mathrm{bbayz}\kern0.24em \mathrm{v}\;(\mathrm{x}=\mathrm{y}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{y})\;\mathrm{v} (\mathrm{x}=\mathrm{z}\;\mathrm{v}\;\mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{z}),$ as required.

(c) That $\mathrm{QT}^{+}\vdash \mathrm{I}_{\leq 2}(\mathrm{x}) \rightarrow \mathrm{x} = \mathrm{a\ v\ aa} \subseteq_\mathrm{p} \mathrm{x}$ is immediate from the definition of $\mathrm{I}^\ast (\mathrm{x})$ .

(d) We show that

$$\begin{align*}\mathrm{QT}^{+} \vdash \mathrm{I}_{\leq 2}(\mathrm{x}) \ \&\ \mathrm{I}_{\leq 2} (\mathrm{y}) \rightarrow ((\mathrm{x} = \mathrm{y\ v\ ax} \subseteq_\mathrm{p} \mathrm{y}) \ \&\ (\mathrm{y} = \mathrm{x\ v\ ay} \subseteq_\mathrm{p} \mathrm{x})\rightarrow \mathrm{x} = \mathrm{y})\end{align*}$$

Assume $\mathrm{M}\models {\mathrm{I}}_{\le 2}(\mathrm{x})\ \&\ {\mathrm{I}}_{\le 2}(\mathrm{y})$ , and suppose, for a reductio, that $\mathrm{M}\models \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{y}\ \&\ \mathrm{ay}{\subseteq}_{\mathrm{p}}\mathrm{x}$ . Then $\mathrm{M}\models \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{y}{\subseteq}_{\mathrm{p}}\mathrm{ay}{\subseteq}_{\mathrm{p}}\mathrm{x}$ . But this contradicts 5.1(g) since $\mathrm{M}\models \mathrm{I}^\ast (\mathrm{x})$ . But then $\mathrm{M}\models (\mathrm{x}=\mathrm{y}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{y})\ \&\ (\mathrm{y}=\mathrm{x}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ay}{\subseteq}_{\mathrm{p}}\mathrm{x})\to \mathrm{x}=\mathrm{y}$ follows.

(e) We show that

$$\begin{align*}&\mathrm{QT}^{+}\vdash \mathrm{I}_{\leq2}(\mathrm{x}) \ \&\ \mathrm{I}_{\leq2} (\mathrm{y}) \ \&\ \mathrm{I}_{\leq2} (\mathrm{z}) \rightarrow\\&\rightarrow ((\mathrm{x} = \mathrm{y\ v\ ax} \subseteq_{\mathrm{p}} \mathrm{y}) \ \&\ (\mathrm{y} = \mathrm{z\ v\ ay} \subseteq_{\mathrm{p}} \mathrm{z}) \rightarrow (\mathrm{x} = \mathrm{z\ v\ ax} \subseteq_{\mathrm{p}} \mathrm{z})).\end{align*}$$

Assume $\mathrm{M}\models {\mathrm{I}}_{\le 2}(\mathrm{x})\ \&\ {\mathrm{I}}_{\le 2}(\mathrm{y})\ \&\ {\mathrm{I}}_{\le 2}(\mathrm{z})$ and suppose $\mathrm{M}\models \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{y}\ \&\ \mathrm{ay}{\subseteq}_{\mathrm{p}}\mathrm{z}$ . Then $\mathrm{M}\models \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{y}{\subseteq}_{\mathrm{p}}\mathrm{ay}{\subseteq}_{\mathrm{p}}\mathrm{z}$ , so $\mathrm{M}\models \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{z}$ . Then $\mathrm{M}\models (\mathrm{x}=\mathrm{y}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{y})\ \& (\mathrm{y}=\mathrm{z}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ay}{\subseteq}_{\mathrm{p}}\mathrm{z})\to (\mathrm{x}=\mathrm{z}\kern0.24em \mathrm{v}\kern0.24em \mathrm{ax}{\subseteq}_{\mathrm{p}}\mathrm{z})$ , as needed.

Proof (a) Assume that $\mathrm{M}\models \mathrm{z}={\tau}_{\mathrm{n}}(\mathrm{x},0,\dots, 0,\mathrm{y})$ where $\mathrm{M}\models \mathrm{I}(\mathrm{x})\ \&\ \mathrm{I}(\mathrm{y})$ . Suppose $\mathrm{M}\models \mathrm{w}\sqsubseteq \mathrm{z}$ where $\mathrm{M}\models \mathrm{w}={\tau}_{\mathrm{n}}({\mathrm{w}}_1,{\mathrm{w}}_2,\dots, {\mathrm{w}}_{\mathrm{n}})$ . We need to show that $\mathrm{M}\models \wedge_{1<\mathrm{i}<\mathrm{n}}\;{\mathrm{w}}_{\mathrm{i}}=0$ . We have that $\mathrm{M}\models \mathrm{w}\sqsubseteq {\tau}_{\mathrm{n}}(\mathrm{x},0,\dots, 0,\mathrm{y})$ . By $(\mathrm{T}{4}_{\mathrm{n}})$ , we get

$$\begin{align*}\mathrm{M}\models \mathrm{w}={\tau}_{\mathrm{n}}(\mathrm{x},0,\dots, 0,\mathrm{y})\kern0.24em \mathrm{v}\;\;\mathrm{w}\sqsubseteq \mathrm{x}\;\;\mathrm{v}\;\;\mathrm{w}\sqsubseteq 0\;\mathrm{v}\dots \mathrm{v}\;\mathrm{w}\sqsubseteq 0\;\;\mathrm{v}\;\;\mathrm{w}\sqsubseteq \mathrm{y}).\end{align*}$$

Proof For (a) and (b), see the proof of 5.1(e)–(g), this time with reference to (3.8) of [Reference Damnjanovic2] for (a) and (3.10) for (b). For (c) and (d), see [Reference Damnjanovic3, 3.7(b) and (c)]. We focus on (e), arguing by induction on $\mathrm{n}$ . Assume that $\mathrm{M} \models \wedge_{1\le \mathrm{i}\le \mathrm{n}}\mathrm{J}^\ast ({\mathrm{x}}_{\mathrm{i}})$ . For $\mathrm{n}=1$ , we have from (d) that

$$\begin{align*}\mathrm{QT}^{+}\vdash \mathrm{wBbax}_1 \rightarrow \mathrm{w} = \mathrm{b\ v\ wBba\ v} = \mathrm{ba\ v}\ \exists \mathrm{z}_1 (\mathrm{z}_1 \mathrm{Bx}_1 \ \&\ \mathrm{w} = \mathrm{baz}_1).\end{align*}$$

Now, $\mathrm{QT}^{+} \vdash \mathrm{wBba} \rightarrow \mathrm{w} = \mathrm{b}$ and $\mathrm{QT}^{+} \vdash \neg \mathrm{wBba} $ , so in fact we have that

$$\begin{align*}\mathrm{QT}^{+}\vdash \mathrm{wBbax}_1 \rightarrow \mathrm{w} = \mathrm{b\ v\ w} = \mathrm{ba\ v}\ \exists \mathrm{z}_1 (\mathrm{z}_1 \mathrm{Bx}_1 \ \&\ \mathrm{w} = \mathrm{baz}_1),\end{align*}$$

as needed. Assume now that the claim holds for $\mathrm{k}$ . We then have that

$$\begin{align*} \begin{array}{l}\mathrm{M}\models {\mathrm{wBb}}^{\mathrm{k}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{k}}\to {\mathrm{wBb}}^{\mathrm{k}}\kern0.24em \mathrm{v}\kern0.24em \mathrm{w}={\mathrm{b}}^{\mathrm{k}}\kern0.24em \mathrm{v}\kern0.24em \mathrm{w}={\mathrm{b}}^{\mathrm{k}}\mathrm{a}\kern0.24em \mathrm{v}\ \exists {\mathrm{z}}_1({\mathrm{z}}_1{\mathrm{Bx}}_1\ \&\ \mathrm{w}={\mathrm{b}}^{\mathrm{k}}{\mathrm{az}}_1)\;\;\mathrm{v}\\ {}\mathrm{v}\kern0.24em {\vee}_{1\le \mathrm{i}<\mathrm{k}}\mathrm{w}={\mathrm{b}}^{\mathrm{k}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{i}}\kern0.24em \mathrm{v}\kern0.24em {\vee}_{1\le \mathrm{i}\le \mathrm{k}}\exists {\mathrm{z}}_{\mathrm{i}}({\mathrm{z}}_{\mathrm{i}}{\mathrm{Bx}}_{\mathrm{i}}\ \&\ \mathrm{w}={\mathrm{b}}^{\mathrm{k}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{i}-1}{\mathrm{z}}_{\mathrm{i}}).\end{array}\end{align*}$$

Assume $\mathrm{M}\models \mathrm{wBb}(({\mathrm{b}}^{\mathrm{k}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{k}}){\mathrm{x}}_{\mathrm{k}+1})$ . By (d),

$$\begin{align*} \mathrm{M}\models \mathrm{w}&=\mathrm{b}\kern0.24em \mathrm{v}\kern0.24em \mathrm{w}\mathrm{Bb}({\mathrm{b}}^{\mathrm{k}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{k}})\;\;\mathrm{v}\kern0.24em \mathrm{w}=\mathrm{b}({\mathrm{b}}^{\mathrm{k}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{k}})\;\mathrm{v}\\ &\qquad\!\!\!\mathrm{v}\;\exists {\mathrm{z}}_{\mathrm{k}+1}({\mathrm{z}}_{\mathrm{k}+1}{\mathrm{Bx}}_{\mathrm{k}+1}\ \&\ \mathrm{w}=\mathrm{b}({\mathrm{b}}^{\mathrm{k}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{k}}){\mathrm{z}}_{\mathrm{k}+1}).\end{align*}$$

From $\mathrm{M}\models \mathrm{w}=\mathrm{b}$ , we have $\mathrm{M}\models \mathrm{wBb}({\mathrm{b}}^{\mathrm{k}})$ , i.e., $\mathrm{M}\models {\mathrm{wBb}}^{\mathrm{k}+1}$ .

Suppose that $\mathrm{M}\models \mathrm{wBb}({\mathrm{b}}^{\mathrm{k}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{k}})$ . By (c), then

$$\begin{align*}\mathrm{M}\models \mathrm{wBb}\kern0.24em \mathrm{v}\kern0.24em \mathrm{w}=\mathrm{b}\kern0.24em \mathrm{v}\;\exists \mathrm{z}({\mathrm{zBb}}^{\mathrm{k}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{k}}\ \&\ \mathrm{bz}=\mathrm{w}).\end{align*}$$

But $\mathrm{QT}^{+}\vdash \neg \mathrm{wBb}$ , and we have $\mathrm{M}\models \mathrm{w}=\mathrm{b}\to {\mathrm{wBb}}^{\mathrm{k}+1}$ . Given the inductive hypothesis, we have from $\mathrm{M}\models {\mathrm{zBb}}^{\mathrm{k}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{k}}\ \&\ \mathrm{bz}=\mathrm{w}$ that

$$\begin{align*} \begin{array}{l}\mathrm{M}\models {\mathrm{z}\mathrm{Bb}}^{\mathrm{k}}\kern0.24em \mathrm{v}\kern0.24em \mathrm{z}={\mathrm{b}}^{\mathrm{k}}\;\mathrm{v}\kern0.24em \mathrm{z}={\mathrm{b}}^{\mathrm{k}}\mathrm{a}\kern0.24em \mathrm{v}\;\exists {\mathrm{z}}_1({\mathrm{z}}_1{\mathrm{Bx}}_1\ \&\ \mathrm{z}={\mathrm{b}}^{\mathrm{k}}{\mathrm{az}}_1)\;\;\mathrm{v}\\ {}\mathrm{v}\kern0.24em {\vee}_{1\le \mathrm{i}<\mathrm{k}}\mathrm{z}={\mathrm{b}}^{\mathrm{k}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{i}}\kern0.24em \mathrm{v}\kern0.24em {\vee}_{1\le \mathrm{i}\le \mathrm{k}}\exists {\mathrm{z}}_{\mathrm{i}}({\mathrm{z}}_{\mathrm{i}}{\mathrm{Bx}}_{\mathrm{i}}\ \&\ \mathrm{z}={\mathrm{b}}^{\mathrm{k}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{i}-1}{\mathrm{z}}_{\mathrm{i}}).\end{array}\end{align*}$$

Now, it is easily seen that $\mathrm{M}\models \mathrm{bz}=\mathrm{w}\ \&\ {\mathrm{zBb}}^{\mathrm{k}}\to {\mathrm{wBb}}^{\mathrm{k}+1}\kern0.24em \mathrm{and}\kern0.24em \mathrm{M}\models \mathrm{bz}=\mathrm{w}\ \&\ \mathrm{z}={\mathrm{b}}^{\mathrm{k}}\to \mathrm{w}={\mathrm{b}}^{\mathrm{k}+1},$ and also that $\mathrm{M}\models \mathrm{bz}=\mathrm{w}\ \&\ \mathrm{z}={\mathrm{b}}^{\mathrm{k}}\mathrm{a}\to \mathrm{w}={\mathrm{b}}^{\mathrm{k}+1}\mathrm{a}.$

Furthermore, $\mathrm{M}\models \mathrm{bz}=\mathrm{w}\ \&\ \exists {\mathrm{z}}_1({\mathrm{z}}_1{\mathrm{Bx}}_1\ \&\ \mathrm{z}={\mathrm{b}}^{\mathrm{k}}{\mathrm{az}}_1)\to \exists {\mathrm{z}}_1({\mathrm{z}}_1{\mathrm{Bx}}_1\ \&\ \mathrm{w} ={\mathrm{b}}^{\mathrm{k}+1}{\mathrm{az}}_1)$ and $\mathrm{M}\models \mathrm{bz}=\mathrm{w}\ \&\ \mathrm{z}={\mathrm{b}}^{\mathrm{k}}{\mathrm{ax}}_1\to \mathrm{w}={\mathrm{b}}^{\mathrm{k}+1}{\mathrm{ax}}_1$ , and so on, plus $\mathrm{M}\models \mathrm{bz}=\mathrm{w}\ \&\ \exists {\mathrm{z}}_{\mathrm{k}}({\mathrm{z}}_{\mathrm{k}}{\mathrm{Bx}}_{\mathrm{k}}\ \&\ \mathrm{z}={\mathrm{b}}^{\mathrm{k}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{k}-1}{\mathrm{z}}_{\mathrm{k}})\to \exists {\mathrm{z}}_{\mathrm{k}}({\mathrm{z}}_{\mathrm{k}}{\mathrm{Bx}}_{\mathrm{k}}\ \&\ \mathrm{w}={\mathrm{b}}^{\mathrm{k}+1} {\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{k}-1}{\mathrm{z}}_{\mathrm{k}}).$

Thus we have

$$\begin{align*} \begin{array}{l}\mathrm{M}\models \mathrm{wBb}({\mathrm{b}}^{\mathrm{k}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{k}})\to {\mathrm{wBb}}^{\mathrm{k}+1}\kern0.24em \mathrm{v}\kern0.24em \mathrm{w}={\mathrm{b}}^{\mathrm{k}+1}\kern0.24em \mathrm{v}\kern0.24em \mathrm{w}={\mathrm{b}}^{\mathrm{k}+1}\mathrm{a}\;\mathrm{v}\\ {}\mathrm{v}\;\exists {\mathrm{z}}_1({\mathrm{z}}_1{\mathrm{Bx}}_1\ \&\ \mathrm{w}={\mathrm{b}}^{\mathrm{k}+1}{\mathrm{az}}_1)\;\;\mathrm{v}\kern0.24em {\vee}_{1\le \mathrm{i}<\mathrm{k}}\mathrm{w}={\mathrm{b}}^{\mathrm{k}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{i}}\;\mathrm{v}\\ {}\mathrm{v}\;{\vee}_{1\le \mathrm{i}\le \mathrm{k}}\exists {\mathrm{z}}_{\mathrm{i}}({\mathrm{z}}_{\mathrm{i}}{\mathrm{Bx}}_{\mathrm{i}}\ \&\ \mathrm{w}={\mathrm{b}}^{\mathrm{k}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{i}-1}{\mathrm{z}}_{\mathrm{i}}).\end{array}\end{align*}$$

But then

$$\begin{align*} \begin{array}{l}\mathrm{M}\models \mathrm{wBb}(({\mathrm{b}}^{\mathrm{k}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{k}}){\mathrm{x}}_{\mathrm{k}+1})\to {\mathrm{wBb}}^{\mathrm{k}+1}\kern0.24em \mathrm{v}\kern0.24em \mathrm{w}={\mathrm{b}}^{\mathrm{k}+1}\kern0.24em \mathrm{v}\kern0.24em \mathrm{w}={\mathrm{b}}^{\mathrm{k}+1}\mathrm{a}\;\mathrm{v}\\ {}\mathrm{v}\;\exists {\mathrm{z}}_1({\mathrm{z}}_1{\mathrm{Bx}}_1\ \&\ \mathrm{w}={\mathrm{b}}^{\mathrm{k}+1}{\mathrm{az}}_1)\;\;\mathrm{v}\kern0.24em {\vee}_{1\le \mathrm{i}<\mathrm{k}+1}\mathrm{w}={\mathrm{b}}^{\mathrm{k}+1}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{i}}\;\mathrm{v}\\ {}\mathrm{v}\kern0.24em {\vee}_{1\le \mathrm{i}\le \mathrm{k}+1}\exists {\mathrm{z}}_{\mathrm{i}}({\mathrm{z}}_{\mathrm{i}}{\mathrm{Bx}}_{\mathrm{i}}\ \&\ \mathrm{w}={\mathrm{b}}^{\mathrm{k}+1}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{i}-1}{\mathrm{z}}_{\mathrm{i}}).\end{array}\end{align*}$$

as required.

Proof (b) Suppose, for a reductio, that $\mathrm{M}\models {\mathrm{b}}^{\mathrm{m}}{\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{m}}={\mathrm{b}}^{\mathrm{n}}{\mathrm{ay}}_1\dots {\mathrm{y}}_{\mathrm{n}}$ where $\mathrm{m}<\mathrm{n}$ . After m applications of $(\mathrm{QT}3)$ we obtain $\mathrm{M}\models {\mathrm{ax}}_1\dots {\mathrm{x}}_{\mathrm{m}}={\mathrm{b}}^{\mathrm{m}-\mathrm{n}}{\mathrm{ay}}_1\dots {\mathrm{y}}_{\mathrm{n}}$ contradicting $(\mathrm{QT}4)$ .

Proof That ${\mathrm{I}}_{\mathrm{Z}}(\mathrm{x})$ is a tree form easily follows from $(\mathrm{T}3)$ and $(\mathrm{T}4)$ . For ${\mathrm{I}}_{\mathrm{REF}\sqsubseteq}(\mathrm{x})$ , note that $\mathrm{T} \vdash \mathrm{I}_{\mathrm{REF} \sqsubseteq} (0)$ follows immediately from $(\mathrm{T}3)$ and $(\mathrm{T}1)$ . Assume $\mathrm{M}\models {\mathrm{I}}_{\mathrm{REF}\sqsubseteq}(\mathrm{u})\ \&\ {\mathrm{I}}_{\mathrm{REF}\sqsubseteq}(\mathrm{v})$ and consider $\mathrm{x}=\tau (\mathrm{u},\mathrm{v})$ . We have that $\mathrm{M}\models \tau (\mathrm{u},\mathrm{v})\sqsubseteq \tau (\mathrm{u},\mathrm{v})$ by $(\mathrm{T}4)$ . On the other hand, assume $\mathrm{M}\models \tau (\mathrm{u},\mathrm{v})=\tau ({\mathrm{x}}_1,{\mathrm{x}}_2)$ . Then $\mathrm{M}\models \mathrm{u}={\mathrm{x}}_1\ \&\ \mathrm{v}={\mathrm{x}}_2$ by $(\mathrm{T}2)$ , whence from $\mathrm{M}\models {\mathrm{I}}_{\mathrm{REF}\sqsubseteq}(\mathrm{u})\ \&\ {\mathrm{I}}_{\mathrm{REF}\sqsubseteq}(\mathrm{v})$ we obtain $\mathrm{M}\models {\mathrm{x}}_1\sqsubseteq {\mathrm{x}}_1\ \&\ {\mathrm{x}}_2\sqsubseteq {\mathrm{x}}_2$ . Hence we have

$$\begin{align*}\mathrm{M}\models \forall {\mathrm{x}}_1,{\mathrm{x}}_2(\tau (\mathrm{u},\mathrm{v})=\tau ({\mathrm{x}}_1,{\mathrm{x}}_2)\to {\mathrm{x}}_1\sqsubseteq {\mathrm{x}}_1\ \&\ {\mathrm{x}}_2\sqsubseteq {\mathrm{x}}_2),\end{align*}$$