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I just need help verifying my answers cause I'm still not 100% what I'm doing at the moment!

Let P and Q be predicates on the set S, where S has two elements, say,$ S = {a, b} $. Then the statement $ ∀xP(x) $ can also be written in full detail as $ P(a) ∧ P(b) $. Rewrite each of the statements below in a similar fashion, using P, Q, and logical operators, but without using quantifiers.

(b) $ ∃xP(x) ∧ ∃xQ(x) $ where I put $ (P(x)∨ P(y)) ∧ (Q(x) ∨ Q(y)) $ . I just was not sure because other example would describe it as $ ∃ x,y P(x) $ making it 2 variable. I just followed the procedure I know, just wanted feedback!

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You need to use $a, b$ to get $$∃xP(x) ∧ ∃xQ(x) \equiv (P(a)∨ P(b)) ∧ (Q(a) ∨ Q(b))$$ since $x \in \{a, b\}$, and can take on either value.

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  • $\begingroup$ well, what would be the difference between what you wrote above and asking making the statement: $ ∃x,yP(x)∧∃xQ(x)≡(P(a)∨P(b))∧(Q(a)∨Q(b) $ ( the difference been x,y) $\endgroup$
    – Marc-Andre Leclair
    Feb 5, 2015 at 18:13
  • $\begingroup$ Thee is no need to make that statement, the $y$ is superfluous, as nothing in the propositions says anything about y. $\endgroup$
    – amWhy
    Feb 5, 2015 at 18:57
  • $\begingroup$ My bad I forgot to do this: $ ∃x,yP(x)∧∃xQ(y)≡(P(a)∨P(b))∧(Q(a)∨Q(b) $ I really don't see the difference between this one ( where Q is Q(y) ) because to me both statement just refer to two different function, so I feel like I can just write the samething for the statement in my initial question and the one I just wrote. $\endgroup$
    – Marc-Andre Leclair
    Feb 5, 2015 at 20:12
  • $\begingroup$ You can write: $∃xP(x)∧∃yQ(y)≡(P(a)∨P(b))∧(Q(a)∨Q(b)$. Logically, it is no different than the statement in the post. $\endgroup$
    – amWhy
    Feb 5, 2015 at 20:15

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