Let $ P$ and $Q$ be predicates on the set $S$, where $S$ has three elements, say, $S = {a, b, c}$. Then the statement $∀xP(x)$ can also be written in full detail as $P(a)∧P(b)∧P(c)$. Rewrite each of the statements below in a similar fashion, using P, Q, and logical operators, but without using quantifiers.
$(b) ∃xP(x)∧∃xQ(x)$
$(c) ∃x, y(P(x)∧Q(y))$
$(d) ∀x∃y(P(x) ∧ Q(y))$
I got the same answer for both part (b) and part (c) and (d), which is
$[P(a) ∨ P(b) ∨ P(c)] ∧ [Q(a) ∨ Q(b) ∨ Q(c)]$
... except it can't be the right answer for all of these statements, i believe. I'm not sure what I did wrong in coming up with these answers.
For (c), think of the possible ways you can assign $(x,y)$ from the set $S$. $P(x) ∧ Q(y)$ must be true for at least one of those -- i.e. you can substitute each pair and join the expressions with a logical 'or'.
(d), from the rule you stated, can be rewritten $$\exists y (P(a) ∧ Q(y)) ∧ \exists y (P(b) ∧ Q(y)) ∧ \exists y (P(c) ∧ Q(y)).$$ From there, it shouldn't be too difficult to replace the $\exists$ with equivalent 'or' statements.