The question i have for an assignment is the following
Let P and Q be predicates on the set S, where S has two elements, say, S = {a, b}. Then the statement ∀xP(x) can also be written in full detail as P(a) ∧ P(b). Rewrite each of the statements below in a similar fashion, using P, Q, and logical operators, but without using quantifiers.
- (a) $\forall x \forall y\, \big(P(x) ∨ Q(y)\big)$
- (b) $\exists x\, P(x) ∧ ∃x\,Q(x)$
- (c) $\exists x\, \exists y\,\big(P(x) ∧ Q(y)\big)$
- (d) $\forall x\,\exists y\,(P(x) ∧ Q(y))$
What throws me off entirely is the x and y, what i would write for a and b:
- (a) $\big(P(a) ∧ P(b)\big) ∨ \big(Q(a) ∧ Q(b)\big)$
- (b) $\big(P(a) ∨ P(b)\big) ∧ \big(Q(a) ∨ Q(b)\big)$
however i know theres something wrong because x and y affect it.... how would i go about this properly?
I'm not sure where you think you are confused. Each time, you have replaced the quantified variables ($x$ and $y$), with the appropriate logical connectives for the predicate for the set of possible values and getting the correct answer.
Or do you not understand how you got your own answers?
To make sure you understand the steps, we split the task up. Do the innermost bound variable first, simplify, then do the outermost bound variable.
$\forall x \forall y\, \big(P(x) ∨ Q(y)\big) \\\equiv \forall x \big((P(x)\vee Q(a))\wedge(P(x)\vee Q(b))\big) \\\equiv \forall x \big(P(x)\vee \big(Q(a)\wedge Q(b)\big)\Big) \\\equiv \Big(P(a)\vee \big(Q(a)\wedge Q(b)\big)\Big)\wedge \Big(P(b)\vee \big(Q(a)\wedge Q(b)\big)\Big) \\\equiv \big(P(a)\wedge P(b)\big)\vee \big(Q(a)\wedge Q(b)\big) $
Which is what you've gotten.
Again, to be sure split the task; do each bound variable one at a time
$\exists x\, P(x) ∧ \exists x\,Q(x) \\\equiv \big(P(a)\vee P(b)\big) \wedge \exists x\, Q(x) \\\equiv \big(P(a)\vee P(b)\big) \wedge \big(Q(a)\vee Q(b)\big)$
Which is almost what you had gotten.Which is what you have (after your edit).