# 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:

\(\displaystyle \) | \(\) | \(\displaystyle x = x\) | Equality is Reflexive | ||||||||||

\(\displaystyle \) | \(\implies\) | \(\displaystyle \exists y: y = x\) | Existential Generalisation | ||||||||||

\(\displaystyle \) | \(\implies\) | \(\displaystyle \left({ \forall y: \left({ y = x \implies P \left({ x }\right) }\right) \implies \exists y: \left({ y = x \land P\left({ x }\right) }\right) }\right)\) | Modus Ponendo Ponens |

$\Box$

\(\displaystyle \left({ y = x \land P \left({ y }\right) }\right)\) | \(\implies\) | \(\displaystyle P \left({x}\right)\) | Substitutivity of Equality | ||||||||||

\((1):\quad\) | \(\displaystyle \implies \ \ \) | \(\displaystyle \exists y: \left({ y = x \land P \left({y}\right) }\right)\) | \(\implies\) | \(\displaystyle P \left({x}\right)\) | Universal Generalisation | ||||||||

\((2):\quad\) | \(\displaystyle \implies \ \ \) | \(\displaystyle \forall y: \left({ y = x \implies P \left({y}\right) }\right)\) | \(\implies\) | \(\displaystyle P \left({x}\right)\) | Hypothetical Syllogism with first lemma |

$\Box$

Similarly:

\(\displaystyle P \left({x}\right)\) | \(\implies\) | \(\displaystyle \left({ y = x \implies P \left({y}\right) }\right)\) | Substitutivity of Equality | ||||||||||

\((3):\quad\) | \(\displaystyle \) | \(\implies\) | \(\displaystyle \forall y: \left({ y = x \implies P \left({y}\right) }\right)\) | Universal Generalisation | |||||||||

\((4):\quad\) | \(\displaystyle \) | \(\implies\) | \(\displaystyle \exists y: \left({ y = x \land P \left({y}\right) }\right)\) | First lemma |

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

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

$\blacksquare$

## Sources

- 1963: Willard Van Orman Quine:
*Set Theory and Its Logic*: $\S 6.1$ - 1963: Willard Van Orman Quine:
*Set Theory and Its Logic*: $\S 6.2$