# Discrete Math. (Logically equivalent)

 P: 562 The question is a little unclear. First, you list four rules of inference for quantifiers in cut form. OK. Then, since you are talking about the all quantifier, I presume you want to refer to the first one. So, you start with $\forall$ x $\in$ D (P(x)$\Rightarrow$ Q(x)) Then, instance with c, P(c)$\Rightarrow$ Q(c). No problem here. But then you say that you can choose any c, including one not in the domain (of the quantifier, I presume you mean, in this case D), which contradicts your original statement which bounded your quantifier to D. Then you start mentioning "E": I am not sure whether you are referring to the existential quantifier $\exists$ or the set membership relation $\in$. Then you say that "it's" not logically equivalent. So far, no equivalence has been mentioned, so what is "it"? and what is it not equivalent to? Please clarify your question, and then I would be glad to help.
 P: 30 ^There were two pages to what he posted. The question asks about: $\exists x ( P(x) \rightarrow Q(x) )$, and $(\forall x) P(x) \rightarrow (\exists x) Q(x)$ Are they logically equivalent? No. There is more than one way to argue it. One obvious thing to take note of is that the in the first statement, the existential binds both instances of little x. This statement can be translated as "there exists an object such that P(x) implies Q(x)" (same x) There is more than one structural difference between the first statement and the second. Can you think of any?
 P: 562 Discrete Math. (Logically equivalent) Before one could even touch the question as to whether the two statements are equivalent, one would have to make sure that they both make sense. The first one, of course, does, but the second one does not even make sense. What is meant, I think, is $\forall$x P(x) $\rightarrow$$\exists$y Q(y). Once that has been cleaned up, one can proceed to show that they are not equivalent.