# ~(~P&Q) & ~(P&Q) : Prove ~Q

This must be proved using only negation, double neg.Intro, double negation Elimination, indirect proof, conj.Intro, and conj.Elim.

• It seems that you know the rules: why do not try to apply them ? – Mauro ALLEGRANZA Nov 30 '18 at 8:39

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:

1. How to get a negative conclusion -- answer assume its opposite
2. How to use indirect proof -- answer find contradictions
• I put "pre" tags around Graham Kemp's formatting. Please feel free to roll back if desired. – Frank Hubeny Dec 1 '18 at 0:33
1. ~(~P&Q) & ~(P&Q) : Prove ~Q

2. (P or ~Q) & ~(P&Q)

3. (P or ~Q) & (~P or ~Q)

4. (P & ~P) or ~Q

5. ~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.

• OP doesn't have access to distribution or demorgan. – virmaior Nov 30 '18 at 6:53