A great deal of the interest in the theory of models of sets of sentences in the lower predicate calculus is due to the increasing success with which this theory has been applied to problems from algebra (see, for example [6, particularly pp. 90–103]).

The existence of nonstandard models prevents this language from being altogether appropriate for application to analysis.

It will be shown later in this paper that the class of all Hilbert spaces is not an elementary class (in the wider sense) in the lower predicate calculus. It is not difficult to find a type and a sentence in Lω1ω1 (of that type) whose models are precisely the Hilbert spaces (slightly altered to include in their domains the real or complex numbers).

It is well known that L, the first-order predicate calculus, is unsuitable for characterizing certain mathematical structures. Since some of those structures are quite often considered by mathematicians, extensions of L have been considered which (partially) solve the above problem. Roughly speaking, the extensions of L fall into two main categories:

(I) extensions obtained by allowing higher order variables, and

(II) extensions obtained by allowing infinitely long formulas.

We are concerned here with the set theory given in [1], which we call BL (Bernays-Levy). This theory can be given an elegant syntactical presentation which allows most of the usual axioms to be deduced from the reflection principle. However, it is more convenient here to take the usual Von Neumann-Bernays set theory [3] as a starting point, and to regard BL as arising from the addition of the schema

where S is the formal definition of satisfaction (with respect to models which are sets) and ┌φ┐ is the Gödel number of φ which has a single free variable X.

§1. In [2], as part of an analogy between the concepts of recursive emimerability and regressiveness, Dekker showed that the intersection of any two regressive sets which are recursively equivalent is a regressive set. In [1], McLaughlin and the author showed that the intersection of two regressive sets, if infinite, has an infinite regressive subset. However, the following theorem shows that the analogy is not complete in this instance.

Theorem. There exist two regressive sets whose intersection is not regressive.

By functional equation we mean an equation of the form

(1) A1 … Aκ = B1 … B1.

Here the A's and B's are functions of one variable from and to the natural numbers and FG is the function obtained from F and G by composition, i.e. FG(x) = F(G(x)) for all natural numbers x. We wish to investigate finite systems of functional equations. Now if all the A's and B's of (1) are equal to the identity function I (or all equal to the zero function O), then the equation (1) is satisfied trivially. Thus, in order to make the problem of solvability of systems of equations interesting, we must have some function given which will be held fixed throughout. We take the successor function S to be this given function.

We wish to make a correction to our paper On definability of ordinals in logic with infinitely long expressions (this journal, vol. 31 (1966), pp. 365–375). Let Ω be an infinite singular cardinal and Ω+ the smallest cardinal >Ω. For the proof of the second half of Theorem 3, we incorrectly assumed that an ordinal γ is not definable in LΩ, if γ is not cofinal with any ordinal ≤Ω. However, Ω+ is indeed definable in LΩ and Theorem 3 should read:

If Ω singular, then an ordinal α is definable in Lα, if and only if σ ≤ exp(Ω+, Ω+), where exp (β,γ) denotes ordinal exponentiation βγ.

In [1] Ludwik Borkowski takes a quantifier symbol ‘Q1’ (e.g., the familiar ‘∀’) to permit definition of another quantifier symbol ‘Q1’ if, where ‘f’ is a singulary predicate variable, there exists a formula A of QC1—a first-order quantificational calculus (without identity and individual constants) having ‘Q1’ as its one primitive quantifier symbol—such that: (1) under the intended interpretations of ‘Q1’ and ‘Q1’ the biconditional (Q1X)f(X) = A is valid, (2) no individual variable occurs free in A, and (3) A contains no propositional variable, and no predicate variable other than ‘f.’

At the Princeton Bicentennial in 1946 Tarski posed the question as to whether or not certain problems connected with partial propositional calculi were recursively solvable. Since that time the specific problems mentioned by Tarski as well as a number of related problems have been shown to be recursively unsolvable. Such results are due to Post and Linial [5], Yntema [9], Gladstone [2], Ihrig [3] and Singletary [7], [8].

As anyone who is familiar with the literature knows, there is a great deal of controversy concerning which, if any, of the extant theories of propositional modal logic correctly formalizes the logic of certain logical concepts such as analyticity and logical necessity. Most of the controversy concerns certain principles that involve iterated modalities (where one modal operator occurs within the scope of another). For example, there is considerable disagreement about whether the principle (□p⊃□□p) should be considered valid. However, when philosophers and logicians apply modal logic to concrete problems, they rarely need principles which involve iterated modalities. For most practical purposes, principles involving only one layer of modalities are all that are needed. This suggests that if we try to construct a theory of modal logic in which there are no iterated modalities, we can avoid most of the controversy and still have a theory that is strong enough for all of the normal uses to which modal logic is put.

We give a short proof of the theorem that, assuming the axiom of choice, every set can be well-ordered. The proof makes direct use of neither induction nor ordinals.

Let S be any given set and its power-set. Let be a choice function, i.e. we denote Ζ — {γΖ} by Ζ'.

Methods are being developed for treating questions of decidability in fewer than Ω steps, Ω being a regular nondenumerable cardinal. In this paper we consider the set-theoretical predicate Taut(x), “x is a tautology,” taken in the infinitary sense. In case x is hereditarily finite there is no question that it is decidable in finitely many steps. But what if x is infinite? We are not assuming that x is in any way constructive or even that the propositional formulas can be well-ordered, so it is not appropriate to treat this predicate as one of natural numbers or even as one of ordinals.