Having trouble proving an L-sentence is not logically valid

Question

I'm having trouble with an assignment in Predicate Calculus specifically related to an L-sentence. I need to prove that

$((\forall x (Px \Rightarrow Qx)) \lor(\forall x(Px \Rightarrow \lnot Qx)))$ is not logically valid.

I understand L-structures and in general how to determine their truth value, however I have gone over this question several times and it looks like it should be logically valid, can you show me how it's not?

(For all x, if Px then Qx) or (For all x, if Px then not Qx) — vfantina, Nov 01 '17 at 21:23
Right. Is it possible for some $a$ to satisfy $\neg(Pa \Rightarrow Qa)$ and for some different $b$ to satisfy $\neg(Pb \Rightarrow \neg Qb)$? — Fabio Somenzi, Nov 01 '17 at 21:26
Sure, if a made Pa True and Qa False, and then b made Pb True and Qb True. Like if Px meant "x is positive" and Qx meant "x is greater than 5" and then we could chose a = 3 and b = 6 — vfantina, Nov 01 '17 at 21:29
Correct. Therefore $(\exists x \neg(Px \Rightarrow Qx)) \wedge (\exists x \neg(Px \Rightarrow \neg Qx))$ is satisfiable. — Fabio Somenzi, Nov 01 '17 at 21:31
Oh my gosh, for some reason I thought the x needed to be the same across the whole equation! So because the whole equation isn't in the scope of the initial $\forall x$ I can have different values, right? — vfantina, Nov 01 '17 at 21:33
Yes, the values of $x$ that contradict the two parts may be different. You could rewrite the second part as $\forall y(Py \Rightarrow \neg Qy)$ without changing its meaning. — Fabio Somenzi, Nov 01 '17 at 21:40

score 0 · Answer 1 · answered Nov 01 '17 at 23:30

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The question was resolved in the comments. Just adding this note so the question stops showing in lists of unanswered questions.

answered Nov 01 '17 at 23:30

Frentos

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Having trouble proving an L-sentence is not logically valid

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