How to show -(P -> -Q) ⊢ P&Q?

Question

I'm unsure how to prove -(P -> -Q) ⊢ P&Q

I know I can assume -(P -> -Q) and use RAA to assume -(P&Q), but from there I'm stuck on how to proceed because of the negations on the outside of the parentheses.

edit: I also keep reading something about DeMorgan's Law, but we haven't learned that yet in class so that's not something I can use (even if it does help in solving this proof).

edit 2: the only rules I know are: assumptions, modus ponendo ponens, modus tollendo tollens, double negation, reductio ad absurdum, conditional proof, v-introduction, v-elimination, &introduction, and &elimination.

Answer 1

1) ¬(P → ¬ Q) --- premise

2) ¬(P & Q) --- assumed [a]

3) P --- assumed [b]

4) Q --- assumed [c]

5) (P & Q) --- from 3) and 4) by &introduction

6) contradicition from 2) and 5)

7) ¬Q --- from 4) and 7) by Indirect Proof, discharging [c]

8) P → ¬ Q --- from 3) and 7) by Conditional Proof, discharging [b]

9) contradicition from 1) and 8)

10) (P & Q) --- from 2) and 9) by Double Negation, discharging [a].

---

Note : for Conditional Proof and Indirect Proof, see e.g. :

• Patrick Hurley, A Concise Introduction to Logic (11th ed - 2012), page 427 and page 432.

Answer 2

As shown elsewhere on this SE, there's two general methods for proofs in sentential logic: truth tables and inference rules. Well for starters, let's just prove exhaustively that the equivalence is true using a truth table:

P | Q | ~Q  | P -> ~Q | ~(P -> ~Q) |  P & Q
--------------------------------------------
T | T |  F  |    F    |   T        |    T
T | F |  T  |    T    |   F        |    F
F | T |  F  |    T    |   F        |    F
F | F |  T  |    T    |   F        |    F

(in this truth table, I've spelled out (a) the variables involved, (b) the steps involved to build ~(P -> ~Q), (c) ~(P -> ~Q) and (d) P & Q. Intermediary steps are a pretty common feature in truth tables especially in classwork).

Given that, we know that the inference is provably, one thing that isn't at all clear from your question is which rules you do know and are allowed to use (you indicate you don't know DeMorgan's or are not allowed to use it).

Here's one proof:

1. | ~(P -> ~Q) A
2. | | ~P A
3. | | ~P v ~Q vI 2
4. | | P -> ~Q Mat. Imp 3
5. | | (P -> ~Q) & ~(P -> ~Q) &I 1,4
6. | P Contra. Elim 2-5
7. | | ~Q A
8. | | ~P v ~Q vI 7
9. | | P -> ~Q Mat. Imp 8
10. | | (P -> ~Q) & ~(P -> ~Q) &I 1,9
11. | Q Contra. Elim 7-10
12. | P & Q &I 6,11
13. ~(P- > ~Q) -> P & Q Conditional Proof 1-12

and there you go.

Edit: here's how to do material implication using conditional proof:

1. | ~(P -> ~Q) A
2. | | ~P A

3. | | ~P v ~Q vI 2

4. | | | P A

5. | | | ~~P DN 4
6. | | | ~Q vE 5,3
7. | | P -> ~Q CP 4-6

... (repeat for the other one and adjust line numbers).

Answer 3

In philosophy I was taught rules of inference. The proof is short as this: 1. ~(p -> ~q). Premise given 2. ~ (~p V ~q) Material implication rule 3. ~ ~ p & ~ ~ q Demorgan's theorem rule 4. P & Q. Double negation rule & conclusion reached.

Solved.