Check this proof please
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Check this proof please

[From: ] [author: ] [Date: 11-10-09] [Hit: ]
y) v Ǝx∀y ¬Q(x,is ¬∀x∀y the same as Ǝx∀y or ƎxƎy?∀x∀y P(x,y) v ƎxƎy ¬Q(x,I think of it this way. pairs of x,......
could someone please tell me if i rewrote it correctly?

¬ ( ƎxƎy ¬P(x,y) ʌ ∀x∀y Q(x,y) )

to

∀xƎy P(x,y) v Ǝx∀y ¬Q(x,y)

is ¬∀x∀y the same as Ǝx∀y or ƎxƎy?

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If I understand this language correctly then it should be

∀x∀y P(x,y) v ƎxƎy ¬Q(x,y)

I think of it this way. pairs of x,y have or dont have certain properties. ƎxƎy ¬P(x,y) means there is a pair (x,y) that doesnt have property P. ∀x∀y Q(x,y) means every pair (x,y) has property Q. Think of x's and y's being real numbers. Property P is that x and y are both positive, property Q is that x and y are both rational. So ( ƎxƎy ¬P(x,y) ʌ ∀x∀y Q(x,y) ) means every pair is rational (Q) and there is a pair that has at least one non-positive element. To negate that we need either every pair has both elements positive (∀x∀y P(x,y) ) OR There is a pair such that one of the elements is not rational ƎxƎy ¬Q(x,y).
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