Past exam of the logic and set theory course of the University of Oxford / 2009 / A / 2

With generalization restricted to variables not free in open assumptions, the deduction theorem statesThe reverse direction follows by thinning and modus ponens. For the forward direction, induct on a derivation, keeping track of its actual assumption contexts. Axioms and assumptions already in $\Gamma$ yield $\alpha \to \chi$ by A1; the distinguished assumption $\alpha$ yields the permitted theorem $\alpha \to \alpha$. If a line $\chi$ is obtained by modus ponens from $\eta$ and $\eta \to \chi$, the induction hypotheses and A2 derive $\alpha \to \chi$.

If a generalization step depends on the discharged assumption, its variable $x$ is not free in $\alpha$ or the other open assumptions. Generalize the induction conclusion $\alpha \to \eta$ to $\forall x(\alpha \to \eta )$ and use A5 to derive $\alpha \to \forall x\eta$. If the subderivation does not depend on $\alpha$, its generalized conclusion can instead be retained as an independent derivation and A1 applied afterwards. Thinning merely enlarges the remaining context and preserves the induction claim. These cases are why assumption dependence, rather than a careless unrestricted universal rule, is essential.

In a presentation allowing generalization freely from hypotheses, the corresponding theorem must instead explicitly require that no variable free in $\alpha$ is generalized in an $\alpha$-dependent line. Without that restriction $P(x)⊢\forall xP(x)$ would incorrectly yield $⊢P(x)\to \forall xP(x)$.

A first-order structure consists of a nonempty domain $A$, an interpretation $c^A\in A$ of each constant symbol, a total function $f^A:A^n\to A$ for each $n$-ary function symbol, and a relation $R^A\subseteq A^n$ for each $n$-ary predicate symbol. Logical equality is interpreted as actual identity.

An assignment in a first-order structure is a map $s$ from the variables into $A$. It extends recursively to terms: $x^A[s]=s(x)$, $c^A[s]=c^A$, and $f(t_1,\dots ,t_n{)}^A[s]=f^A(t_1^A[s],\dots ,t_n^A[s])$. Atomic truth is determined by membership of the resulting tuple in $R^A$, or equality of the two term values. Negation reverses the truth value, and an implication is false exactly when its antecedent is true and its consequent false.

For a universal quantifier,These clauses define a valuation on every formula by structural recursion. Only free-variable values affect a formula's truth; consequently a sentence has the same truth value under every assignment. This is a semantic assignment of domain elements, not a programming assignment operation.

Fix any structure and any assignment of the three quantified variables, writing their values as $a_0,a_1,a_2$. If either of the nested antecedents is false, the corresponding implication is true. If both are true, then $f^A(a_0)=a_1$ and $f^A(a_0)=a_2$. Actual identity is transitive, so $a_1=a_2$, making the consequent true as well.

Thus the unquantified matrix is true under every assignment in every structure. The universal-quantifier semantic clause then makes its universal closure true in every structure:No assumption that $f$ is injective is used. The argument says that one value cannot equal two distinct elements, not that distinct inputs have distinct outputs.

Choose a fresh variable $x_3$. In A7 take the equality antecedent $x_3=x_1$, the atomic formula $\varphi =(x_3=x_2)$, and replace its occurrence of $x_3$ by $x_1$. This gives directly the theoremGeneralize on $x_3$, with no open assumptions. A4 then instantiates that quantifier with the term $f(x_0)$. The substitution is free for $x_3$, since the matrix contains no quantifiers. One modus-ponens step yieldsFinally generalize, in order, on $x_2$, $x_1$, and $x_0$. The result isEvery step is A7, generalization, A4, or modus ponens. Neither completeness, adequacy, nor an unproved appeal to all propositional tautologies is needed. In particular the proof does not silently upgrade the variable-only equality axiom to an arbitrary-term axiom: quantification and instantiation perform that upgrade explicitly.