Taylor-Grey M. answered • 06/04/19

Philosophy PhD candidate working on logic

For the first:

We know that, universal quantifier distribute over conjunction so:

Ax [ ~P(x) ^ ~ Q(x)] if and only if [Ax ~P(x)] ^ [Ax ~Q(x)]

Negate both sides

~Ax [~P(x) ^ ~Q(x)] if and only if ~([~Ax P(x)] ^ [Ax ~Q(x)])

Quantifier Negate

Ex~[~ P(x) ^ ~Q(x)] if and only if ~ ([Ax ~ P(x)] ^ [Ax ~ Q(x)])

DeMorgan

Ex [P(x) v Q(x)] if and only if ~ ([Ax ~ P(x)] ^ [Ax ~ Q(x)])

Quanitifer Negate and Double Negation

Ex [P(x) v Q(x)] if and only if [Ex P(x)] v [Ex Q(x)]

then you can do biconditional exchange to get

Ex[P(x) v Q(x)] only if [Ex P(x)] v [Ex Q(x)]

if you have to include the Ex[P(x) v Q(x)] as premise 1 then do everything above starting on line 2.

When you get Ex[P(x) v Q(x)] only if [Ex P(x)] v [Ex Q(x)] you'll be able to apply modus ponens.

For the second one the proof could look like this

- Ex(Px ^ Qx)
- Pa ^ Qa (1)By EI (existential instantiation
- Pa (2)By conjunction elimination
- Qa (2)By conjunction elimination
- ExPx (3) by EG (existential generalization)
- ExQx (3) by EG
- ExPx ^ ExQx (5) and (6) by conjunction introduction