Prove: [vdash forall x (P(x) rightarrow Q(x)) rightarrow (exists x P(x) rightarrow exists x Q(x))]

The statement being proven is "For all x, if P(x) implies Q(x), then if there exists an x such that P(x) is true, then there exists an x such that Q(x) is true." This can be proven by assuming the antecedent, \forall x (P(x) \rightarrow Q(x)), and showing that if \exists x P(x) is true, then \exists x Q(x) must also be true. Essentially, this proof demonstrates that if every instance of P implies Q, then if there is at least one case where P is true, there must also be at least one case where Q is true.

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