Discussion
Let P(x, y) be a propositional function, then .................. is a tautology.
Answer: C
The tautology is ∃x∀y P(x, y) → ∀y∃x P(x, y).
If some single x0 makes P(x0, y) true for every y, then for each y there is certainly an x making P(x, y) true, namely that same x0. So the implication holds for every predicate P and every domain: it is valid.
The others can fail. In (A) and (B), knowing that P holds for some x at one fixed y0 says nothing about P at other values of y. In (D), ∃x∀y P(x, y) guarantees some x that works for all y, but not that the particular fixed value x0 is that x. The converse of the correct option, ∀y∃x → ∃x∀y, is also not valid, which is the usual trap.
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