Substitutivity of Class Equality

From ProofWiki
Jump to navigation Jump to search


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.

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



This page is beyond the scope of ZFC, and should not be used in anything other than the theory in which it resides.

If you see any proofs that link to this page, please insert this template at the top.

If you believe that the contents of this page can be reworked to allow ZFC, then you can discuss it at the talk page.

By induction on the well-formed parts of $P \left({A}\right)$.

The proof shall use $\implies$ and $\neg$ as the primitive connectives.


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:

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


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

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 $P$


Also see