In your second formula, x is not even bounded in the second part, so $$(∀x~ fsa(x))→(∃y~ pda(y)∧equivalent(x,y))$$ contains a free variable.
Even considering its universal closure, it would not mean what you want.
In your second formula, x is not even bounded in the second part, so $$(∀x~ fsa(x))→(∃y~ pda(y)∧equivalent(x,y))$$ contains a free variable.
Even considering its universal closure, it would not mean what you want.