# logic – Is \$P(a)\$ logically equivalent to \$forall y [(y=a) rightarrow P(y)]\$?

I am not quite sure if my proof is entirely correct. Help is much appreciated.

$$(rightarrow)$$
Assume $$P(a)$$. Let an arbitrary $$y$$. Let $$y=a$$. Since $$P(a)$$ and $$y=a$$, then $$P(y)$$. Since $$y$$ is arbitrary, then $$forall y ((y=a) rightarrow P(y))$$.

$$(leftarrow)$$ Assume $$forall y ((y=a) rightarrow P(y))$$. Then, by universal instantiation, $$(a=a) rightarrow P(a)$$. Then, $$P(a)$$.

Since $$P(a)$$ is logically equivalent to $$forall y ((y=a) rightarrow P(y))$$, then, assuming $$Gamma$$ is a set of formulas, $$Gamma rightarrow P(a)$$ is equivalent to $$Gamma rightarrow forall y ((y=a) rightarrow P(y))$$. If $$y$$ does not occur in $$Gamma$$, then the statement is equivalent to $$forall y (Gamma rightarrow ((y=a) rightarrow P(y)))$$, which is the same as $$forall y ((Gamma land (y=a)) rightarrow P(y))$$. Have I missed something?