This must be proved using only negation, double neg.Intro, double negation Elimination, indirect proof, conj.Intro, and conj.Elim.
By hand, I get:
1. |_ ~(~P&Q) & ~(P&Q) A 2. | ~(~P&Q) &E 1 3. | ~(P&Q) &E 1 4. | |_ Q A 5. | | | P A 6. | | | P&Q &I 4,5 7. | | | ~(P&Q) & (P&Q) &I 3,6 8. | | ~P IP 5-7 9. | | ~P & Q &I 4,8 10. | | (~P&Q) & ~(~P&Q) &I 9,2 11. | ~Q IP 4-10
Conceptually, what we need to think about is:
- How to get a negative conclusion -- answer assume its opposite
- How to use indirect proof -- answer find contradictions
~(~P&Q) & ~(P&Q) : Prove ~Q
(P or ~Q) & ~(P&Q)
(P or ~Q) & (~P or ~Q)
(P & ~P) or ~Q
~Q (By Contradiction)
In step 2 we distribute the negation inside the parenthesis and use de morgan's laws. Same for step 3 for the other parenthesis with negation. In step 4 we use distributive law to put ~Q outside of the and. In step 5 we can prove ~Q by contradiction. You can expand this out if you need to. The left side of the or statement in 4. will always be false (true & false === false). Thus the right statement must be true for the logical statement to be true. Thus ~Q.
When I first learned logic we didn't use any of the terms you used in your post- hopefully nothing in my proof is deemed invalid for you to use. At the very least you can get a general idea of one proof for your statement.