PREDICATE LOGIC

Question

Consider the following predicate statements.

P1: ~$\forall$x ~(P(X)-->$\exists$yQ(y))

P2: $\exists$x(~P(x)$\vee$ $\exists$yQ(y))

P3: $\exists$x(~$\exists$yQ(y)-->~P(x))

P4: ~$\forall$x(P(X) $\wedge$ ~$\exists$Q(Y))

which of the above predicates are equivalent to the predicate statement:

                      $\exists$x(P(x)$\rightarrow$ $\exists$Q(y))

A)P1,P2,P3  B)P1,P3,P2 C)P2,P3,P4  D)ALL OF THESE.

P2 is trivial,but in others predicate i am not able to see how the scope of ~(negation) changes,with respect to parantheses.. 

Comments

all are equivalent ...

option1) operate outer most ~ which give same expresion

option 2) its simplification of implication

option 3 its contrapositive of given expression so its equivalent to implication

option 4again operate ~ which gives you 2nd expression

Answer 1

P1: ~∀x ~(P(X)-->∃yQ(y)) = ∃x~~(P(X)-->∃yQ(y)) = ∃x(P(X)-->∃yQ(y))

P2: ∃x(~P(x)∨ ∃yQ(y)) =∃x(P(x)--> ∃yQ(y))    {p-->q = ~p v q}

P3: ∃x(~∃yQ(y)-->~P(x))  = ∃x(~~∃yQ(y) v ~P(x)) =∃x(∃yQ(y) v ~P(x)) =  ∃x(P(X)-->∃yQ(y))

P4: ~∀x(P(X) ∧ ~∃Q(Y)) = ∃x(~P(X) v ~~∃Q(Y)) = ∃x(~P(X) v ∃Q(Y)) = ∃x(P(X)-->∃yQ(y))

Hence, all are equivalent.