Removing negations using logical equivalences

[idlackage]

Homework Statement

Remove negations from this: ¬∀x∃y : (P (x, y) --> (∃z : ¬Q(x, y, z))) using logical equivalences.

Homework Equations

¬(∃x : P(x)) is equal to ∀x : ¬(P(x))
P --> Q is equal to ¬P v Q

The Attempt at a Solution

¬∀x∃y : (P (x, y) --> (∃z : ¬Q(x, y, z)))
≡ ¬∀x∃y : (¬P (x, y) v (∃z : ¬Q(x, y, z)))
≡ ∃x∀y : ¬ (¬P (x, y) v (∃z : ¬Q(x, y, z)))
≡ ∃x∀y : (P (x, y) v ¬(¬(∀z : Q(x, y, z))))
≡ ∃x∀y : (P (x, y) v (∀z : Q(x, y, z)))

I'm wondering if I've applied the logic right. I'm not sure if I can just shift the Not like that from line one to two (counting from the equivalency sign). On line two to line three, I'm assuming that (∃z : ¬Q(x, y, z)) would mean something like "There exists a person who doesn't like math", which would equal to ¬(∀z : Q(x, y, z)) or "Not everyone likes math". Then on line four the Not from the very outside cancels the Not in front of that ∀z. Am I doing this right? I feel like I'm missing something.

Thanks.

 

[KataKoniK]

Your approach is generally correct, but there are a few small errors. Here is a slightly modified version of your solution:

¬∀x∃y : (P (x, y) --> (∃z : ¬Q(x, y, z)))
≡ ¬∃x∀y : (P (x, y) --> (∃z : ¬Q(x, y, z))) (using the fact that ¬(∃x : P(x)) is equal to ∀x : ¬(P(x)))
≡ ∃x∀y : ¬(P (x, y) --> (∃z : ¬Q(x, y, z))) (using the fact that P --> Q is equal to ¬P v Q)
≡ ∃x∀y : (P (x, y) v ¬(∃z : ¬Q(x, y, z))) (using the fact that ¬(P --> Q) is equal to P v ¬Q)
≡ ∃x∀y : (P (x, y) v (∀z : Q(x, y, z))) (using the fact that ¬(∃z : ¬Q(x, y, z)) is equal to ∀z : Q(x, y, z))

So the main error in your solution was on line 3, where you incorrectly applied the distributive property of negation. You can't just "shift the Not" from one part of the expression to another - you have to use the logical equivalences to rewrite the entire expression. Other than that, your logic is correct. Good job!