Substitutivity of Class Equality

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Theorem

Let $A$ and $B$ be classes.

Let $P \left({A}\right)$ be a well-formed formula of the language of set theory.

Let $P \left({B}\right)$ be the same proposition $P \left({A}\right)$ with all instances of $A$ replaced with instances of $B$.

Let $=$ denote class equality.


$A = B \implies \left({ P \left({A}\right) \iff P \left({B}\right) }\right)$


Proof

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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 $C$:

\(\displaystyle A = B\) \(\implies\) \(\displaystyle \left({ x = A \iff x = B }\right)\) Class Equality is Transitive
\(\displaystyle \) \(\implies\) \(\displaystyle \left({ \left({ x = A \land x \in C }\right) \iff \left({ x = B \land x \in C }\right) }\right)\) Propositional Manipulation
\(\displaystyle \) \(\implies\) \(\displaystyle \forall x: \left({ \left({ x = A \land x \in C }\right) \iff \left({ x = B \land x \in C }\right) }\right)\) Universal Generalisation
\(\displaystyle \) \(\implies\) \(\displaystyle \left({ A \in C \iff B \in C }\right)\) Definition of Class Membership

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

\(\displaystyle A = B\) \(\implies\) \(\displaystyle \left({ x \in A \iff x \in B }\right)\) Definition of Class Equality
\(\displaystyle \) \(\implies\) \(\displaystyle \left({ \left({ x = C \land x \in A }\right) \iff \left({ x = C \land x \in B }\right) }\right)\) Propositional Manipulation
\(\displaystyle \) \(\implies\) \(\displaystyle \forall x: \left({ \left({ x = C \land x \in A }\right) \iff \left({ x = C \land x \in B }\right) }\right)\) Universal Generalisation
\(\displaystyle \) \(\implies\) \(\displaystyle \left({ C \in A \iff C \in B }\right)\) Definition of Class Membership


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:

\(\displaystyle A = B\) \(\implies\) \(\displaystyle \left({ Q \left({x,z}\right) \iff Q \left({y, z}\right) }\right)\) Inductive Hypothesis
\(\displaystyle \) \(\implies\) \(\displaystyle \forall z: \left({ Q \left({x,z}\right) \iff Q \left({y, z}\right) }\right)\) Universal Generalization
\(\displaystyle \) \(\implies\) \(\displaystyle \left({ \forall z: Q \left({x, z}\right) \iff \forall z: Q \left({y, z}\right) }\right)\) Predicate Logic manipulation
\(\displaystyle \) \(\implies\) \(\displaystyle \left({ P \left({A}\right) \iff P \left({B}\right) }\right)\) Definition of $P$

$\blacksquare$


Also see


Sources