Equality implies Substitution

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Theorem

Let $P \left({x}\right)$ denote a Well-Formed Formula which contains $x$ as a free variable.

Then the following are tautologies:

$\forall x: \left({P \left({ x }\right) \iff \exists y: \left({y = x \land P \left({y}\right)}\right)}\right)$
$\forall x: \left({P \left({ x }\right) \iff \forall y: \left({y = x \implies P \left({y}\right)}\right)}\right)$

Note that when $y$ is substituted for $x$ in either formula, it is false in general; compare Confusion of Bound Variables.


Proof

By Universal Affirmative implies Particular Affirmative iff First Predicate is not Vacuous:

$\left({\exists y: y = x \land \forall y: \left({y = x \implies P \left({x}\right)}\right)}\right) \implies \exists y: \left({y = x \land P \left({x}\right)}\right)$

Then:

\(\displaystyle \) \(\) \(\displaystyle x = x\) $\quad$ Equality is Reflexive $\quad$
\(\displaystyle \) \(\implies\) \(\displaystyle \exists y: y = x\) $\quad$ Existential Generalisation $\quad$
\(\displaystyle \) \(\implies\) \(\displaystyle \left({ \forall y: \left({ y = x \implies P \left({ x }\right) }\right) \implies \exists y: \left({ y = x \land P\left({ x }\right) }\right) }\right)\) $\quad$ Modus Ponendo Ponens $\quad$

$\Box$


\(\displaystyle \left({ y = x \land P \left({ y }\right) }\right)\) \(\implies\) \(\displaystyle P \left({x}\right)\) $\quad$ Substitutivity of Equality $\quad$
\((1):\quad\) \(\displaystyle \implies \ \ \) \(\displaystyle \exists y: \left({ y = x \land P \left({y}\right) }\right)\) \(\implies\) \(\displaystyle P \left({x}\right)\) $\quad$ Universal Generalisation $\quad$
\((2):\quad\) \(\displaystyle \implies \ \ \) \(\displaystyle \forall y: \left({ y = x \implies P \left({y}\right) }\right)\) \(\implies\) \(\displaystyle P \left({x}\right)\) $\quad$ Hypothetical Syllogism with first lemma $\quad$


$\Box$


Similarly:

\(\displaystyle P \left({x}\right)\) \(\implies\) \(\displaystyle \left({ y = x \implies P \left({y}\right) }\right)\) $\quad$ Substitutivity of Equality $\quad$
\((3):\quad\) \(\displaystyle \) \(\implies\) \(\displaystyle \forall y: \left({ y = x \implies P \left({y}\right) }\right)\) $\quad$ Universal Generalisation $\quad$
\((4):\quad\) \(\displaystyle \) \(\implies\) \(\displaystyle \exists y: \left({ y = x \land P \left({y}\right) }\right)\) $\quad$ First lemma $\quad$

The above two statements comprise the other direction of the biconditional assertions.

Together, $(1)$, $(2)$, $(3)$, and $(4)$ prove the two assertions.

$\blacksquare$


Sources