Evaluate the truth value of the formula (∃x)(P(x)∧Q(x)) ⟹ ((∃xP(x))∧(∃xQ(x)))(\exists x)(P(x) \land Q(x)) \implies ((\exists x P(x)) \land (\exists x Q(x)))(∃x)(P(x)∧Q(x))⟹((∃xP(x))∧(∃xQ(x))) in First-Order Logic.
It is a tautology.
It is a contradiction.
It is satisfiable but not a tautology.
It is false for all domains.