Past exam of the logic and set theory course of the University of Oxford / 2011 / A / 3

A logical derivation is a finite proof in the specified calculus, with conclusion $\Gamma ⊢\varphi$. Assumptions from $\Gamma$ and instances of A1–A7 are available; the rules are the modus ponens rule, universal generalization rule, and weakening.

In judgement form the universal generalization rule takes $\Delta ⊢\theta$ to $\Delta ⊢\forall x_i\theta$ only if $x_i$ is not free in any undischarged assumption in $\Delta$. Weakening permits $\Delta ⊢\theta$ to become $\Delta \cup \Sigma ⊢\theta$. All instances of A4 and A5 must meet their stated substitution and free variable conditions. Derivability means existence of such a finite proof with assumptions contained in $\Gamma$. Equivalently one tracks the finite assumption set used by each proof line.

The first-order term $t$ is free for substitution for $x_i$ in $\varphi$ if, at every free occurrence of $x_i$ to be replaced, no enclosing quantifier binds a variable occurring in $t$. Equivalently, the substitution introduces no variable capture. Only free occurrences of $x_i$ are replaced.

Take $\varphi =\forall x_1(x_0=x_1)$. The term $x_2$ is free for substitution for $x_0$: its variable is not bound by the enclosing quantifier. The term $x_1$ is not free for substitution for $x_0$, because insertion under $\forall x_1$ captures that occurrence.

Thus the first substitution gives $\forall x_1(x_2=x_1)$; the second, illegitimate substitution would give $\forall x_1(x_1=x_1)$.

In a two-element first-order structure, $\varphi =\forall x_1(x_0=x_1)$ is false for every first-order variable assignment. Blind substitution of $x_1$ for $x_0$ yields $\forall x_1(x_1=x_1)$, which is true. This does not express the original property evaluated at the value of the inserted term: it has changed a free parameter into a bound variable.

For a direct failure of the substitution axiom, use $\theta (x_0)=\exists x_1(x_0\ne x_1)$. On a two-element domain, $\forall x_0\theta (x_0)$ is true, whereas the captured instance $\exists x_1(x_1\ne x_1)$ is false. Consequently an unrestricted A4 would prove a false implication. The free for substitution condition is necessary for soundness.

Write $E(x,y)$ for the displayed equality predicate. A6 and A4 give $⊢E(x,x)$, establishing a reflexive relation.

An instance of A7, replacing just the first occurrence of $x$ in $E(x,x)$, isAssume $E(x,y)$, apply the modus ponens rule twice using reflexivity, and obtain $E(y,x)$. The deduction theorem for propositional logic discharges the assumption. This proves symmetry.

For transitivity, A7 suppliesby replacing $y$ in the atomic formula $E(x,y)$. From assumptions $E(x,y)$ and $E(y,z)$, two applications of the modus ponens rule give $E(x,z)$. Discharging assumptions and universally generalizing givesThus the predicate defines an equivalence relation within the calculus.

No such first-order axiom collection can force actual identity when the displayed symbol is treated as an ordinary relation. The duplication obstruction to axiomatizing identity proves this even for an infinite collection of extra axioms.

Start with any nonempty first-order structure $\mathcal{A}$ in which the symbol is identity. Form $B=A\times \{0,1\}$ and the surjection $\pi :B\to A$. Interpret each relation by pulling it back along $\pi$; in particular,Interpret every function on first coordinates as in $\mathcal{A}$ and choose second coordinate zero; likewise lift constants with second coordinate zero. Induction on first-order terms gives $\pi (t^{\mathcal{B}}(v))=t^{\mathcal{A}}(\pi v)$. Induction on first-order formulas then givesThe quantifier steps use surjectivity of $\pi$, so every witness in $A$ lifts to $B$.

Every proposed axiom valid with actual equality is therefore true in $\mathcal{B}$ under every assignment, along with A1–A7. But $E^{\mathcal{B}}((a,0),(a,1))$ holds for two different domain elements. Thus those axioms cannot enforce identity. The usual interpretation of logical equality as identity is a semantic requirement.