(a) ∀x∃y(P(x) ∧ Q(y)) ⇔ ∃x∀y(P(x) ∧ Q(y)) (b) ∀x∃y(P(x) → Q(y)) ⇔ ∃x∀y(P(x) → Q(y)) (c) ¬∀x(P(x) ∧ Q(x)) ⇔ ∃y(P(y) → ¬Q(y))
So with these questions im supposed to show that in each case the equivalency is true no matter what sets x and y are apart of. I can do that or show a counterargument. I need help badly if anyone knows how to solve these please help. Thanks in advance
The LHS of (a) says that 'for every $x$ there is some $y$ such that $x$ has property $P$ and $y$ has property $y$. Note that this implies that everything needs to have property $P$, but you really need only one thing to have property $Q$.
The RHS, however, says the 'opposite': for the RHS to be true everything needs to have property $Q$, but only one thing needs to have property $P$.
So, these are not saying the same thing ... and for a concrete counterexample, you just need two objects: if both objects have property $P$ but only one of them has property $Q$, then the LHS is true, but the RHS is false.
OK, can you try and do (b) and (c) like that?