How do one justify $( \exists x) (\forall y) \, P(y) \equiv \forall y \, P(y) $. ? I came across a proof using this justification and i am unclear. Thanks
Context:
$\tag 1 \exists x \; [P(x) \Rightarrow \forall y \, P(y)]$
Claim: (1) is a tautology, since it leads to an equivalence chain that ends in a well-known tautology:
$\tag 2 \exists x \; [ \neg P(x) \, \text{ or } \, \forall y \, P(y)]$
$\tag 3 \exists x \; [ \neg P(x)] \, \text{ or } \, \forall y \, P(y)$
$\tag 4 \neg \, \forall x \, P(x) \, \text{ or } \, \forall y \, P(y)$
$\tag 5 \neg \, [\forall z \, P(z)] \, \text{ or } \, [\forall z \, P(z)]$
The equivalence chain ends with a tautology, so (1) must also be one.
We can go from (2) to (3) since the existential qualifier distributes over disjunction and $( \exists x) (\forall y) \, P(y) \equiv \forall y \, P(y) $.
