¬∀xP (x) ≡ ∃x¬P(x)

how could we show that ?

if I said:
¬∀xP (x) is true , if and only if ∀xP (x) is false , then there's p(x) where ¬∀xP (x ) is true and ∀xP (x) is false.
and ∃x¬P(x) is true if and only if ∃x P(x) is false it follows that ¬P(x) is true in case there's at least an x where ¬P(x) is true , and P(x) is false in case there's no x that would make the statement true.

would this answer work ? or how could we prove ?

Question

¬∀xP (x) ≡ ∃x¬P(x)

how could we show that ? 

if I said:
¬∀xP (x) is true , if and only if ∀xP (x) is false , then there's p(x) where ¬∀xP (x ) is true and ∀xP (x) is false.
and ∃x¬P(x) is true if and only if ∃x P(x) is false it follows that ¬P(x) is true in case there's at least an x where ¬P(x) is true , and P(x) is false in case there's no x that would make the statement true.

would this answer work ? or how could we prove ?

amistre64 · Answer

Not all of x are in P(x) is equivalent to saying, there exists an x that is not in P(x)

hmm

amistre64 · Answer

im with heaviside ... it seems rather pointless to try to prove that what you say is what you say :/

amistre64 · Answer

Let x be an element of X that is not in P(x).  Therefore there exists an x in X that is in notP(x) ???

amistre64 · Answer

i got no good ideas for this one :/

anonymous · Answer

:( oh man it's kinda hard I know

I honestly hate such questions , I think i'll leave it there and continue studying and get to this question after I finish.

I'll let this question open if anyone got an answer for it.