Substitutivity of Class Equality

Theorem
Let $A$ and $B$ be classes.

Let $\map P A$ be a well-formed formula of the language of set theory.

Let $\map P B$ be the same proposition $\map P A$ with all instances of $A$ replaced with instances of $B$.

Let $=$ denote class equality.


 * $A = B \implies \paren {\map P A \iff \map P B}$

Proof
By induction on the well-formed parts of $\map P A$.

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 $C$:

Then, if $A$ and $B$ have $C$ as a member:

Inductive Step for $\implies$
Suppose that $\map P A$ is of the form $\map Q A \implies \map R A$

Furthermore, suppose that:
 * $A = B \implies \paren {\map Q A \iff \map Q B}$

and:
 * $A = B \implies \paren {\map R A \iff \map R B}$

It follows that:


 * $A = B \implies \paren {\paren {\map Q A \implies \map R A} \iff \paren {\map Q B \implies \map R B} }$


 * $A = B \implies \paren {\map P A \iff \map P B}$

Inductive Step for $\neg$
Suppose that $\map P A$ is of the form $\neg \map Q A$

Furthermore, suppose that:


 * $A = B \implies \paren {\map Q A \iff \map Q B}$

It follows that:


 * $A = B \implies \paren {\neg \map Q A \iff \neg \map Q B}$


 * $A = B \implies \paren {\map P A \iff \map P B}$

Inductive Step for $\forall x:$
Suppose that $\map P A$ is of the form $\forall z: \map Q {x, z}$

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:

Also see

 * Axiom:Leibniz's Law