{{NotZFC}} By induction on the well-formed parts of $P \left({A}\right)$. The proof shall use $\implies$ and $\neg$ as the primitive connectives. === Atoms === First, to prove the statement if $A$ and $B$ are members of some other class $C$: {{begin-eqn}} {{eqn | l = A = B       | o = \implies       | r = \left({ x = A \iff x = B }\right)       | c = Class Equality is Transitive }} {{eqn | o = \implies       | r = \left({ \left({ x = A \land x \in C }\right) \iff \left({ x = B \land x \in C }\right) }\right)       | c = Propositional Manipulation }} {{eqn | o = \implies       | r = \forall x: \left({ \left({ x = A \land x \in C }\right) \iff \left({ x = B \land x \in C }\right) }\right)       | c = Universal Generalisation }} {{eqn | o = \implies       | r = \left({ A \in C \iff B \in C }\right)       | c = Definition of Class Membership }} {{end-eqn}} Then, if $A$ and $B$ have $C$ as a member: {{begin-eqn}} {{eqn | l = A = B       | o = \implies       | r = \left({ x \in A \iff x \in B }\right)       | c = Definition of Class Equality }} {{eqn | o = \implies       | r = \left({ \left({ x = C \land x \in A }\right) \iff \left({ x = C \land x \in B }\right) }\right)       | c = Propositional Manipulation }} {{eqn | o = \implies       | r = \forall x: \left({ \left({ x = C \land x \in A }\right) \iff \left({ x = C \land x \in B }\right) }\right)       | c = Universal Generalisation }} {{eqn | o = \implies       | r = \left({ C \in A \iff C \in B }\right)       | c = Definition of Class Membership }} {{end-eqn}} === Inductive Step for $\implies$ === Suppose that $P \left({A}\right)$ is of the form $Q \left({A}\right) \implies R \left({A}\right)$ Furthermore, suppose that: :$A = B \implies \left({ Q\left({A}\right) \iff Q \left({B}\right) }\right)$ and: :$A = B \implies \left({ R\left({A}\right) \iff R \left({B}\right) }\right)$ It follows that: :$A = B \implies \left({ \left({ Q \left({A}\right) \implies R \left({A}\right) }\right) \iff \left({ Q \left({B}\right) \implies R \left({B}\right) }\right) }\right)$ :$A = B \implies \left({ P \left({A}\right) \iff P \left({B}\right) }\right)$ === Inductive Step for $\neg$ === Suppose that $P \left({A}\right)$ is of the form $\neg Q \left({A}\right)$ Furthermore, suppose that: :$A = B \implies \left({ Q \left({A}\right) \iff Q \left({B}\right) }\right)$ It follows that: :$A = B \implies \left({ \neg Q \left({A}\right) \iff \neg Q \left({B}\right) }\right)$ :$A = B \implies \left({ P \left({A}\right) \iff P \left({B}\right) }\right)$ === Inductive Step for $\forall x:$ === Suppose that $P \left({A}\right)$ is of the form $\forall z: Q \left({x, z}\right)$ If $x$ and $z$ are the same variable, then $x$ is a bound variable and the theorem holds trivially. If $x$ and $z$ are distinct, then: {{begin-eqn}} {{eqn | l = A = B       | o = \implies       | r = \left({ Q \left({x,z}\right) \iff Q \left({y, z}\right) }\right)       | c = Inductive Hypothesis }} {{eqn | o = \implies       | r = \forall z: \left({ Q \left({x,z}\right) \iff Q \left({y, z}\right) }\right)       | c = Universal Generalization }} {{eqn | o = \implies       | r = \left({ \forall z: Q \left({x, z}\right) \iff \forall z: Q \left({y, z}\right) }\right)       | c = Predicate Logic manipulation }} {{eqn | o = \implies       | r = \left({ P \left({A}\right) \iff P \left({B}\right) }\right)       | c = Definition of $P$ }} {{end-eqn}} {{qed}}