1.) ∃x(P(x)->Q(x)) ->(∀xP(x)->∀xQ(x))
let P(x)= x is natural no
let Q(x)= x is odd no
let x is the set of natural no
to make above statement false
we should make RHS false and and LHS true
RHS is (∀xP(x)->∀xQ(x)) is equivalent to (T--> F) is false
LHS is ∃x(P(x)->Q(x)) is equivalent to true
so ∃x(P(x)->Q(x)) ->(∀xP(x)->∀xQ(x)) is false
2.) ∃xP(x)->∀x Q(x) ->∀x(P(x)->Q(x))
let P(x)= x is even no
let Q(x)= x is odd no
let x is set of natural no
RHS is ∀x(P(x)->Q(x)) equivalent to false
LHS is ∃xP(x)->∀x Q(x) equivalent to (T-->F) is false
so ∃xP(x)->∀x Q(x) ->∀x(P(x)->Q(x)) is true
not any case to make it false