Here is my attempt, but I'm really not sure if I've done it right; as I'm just about getting the hang of Natural Deduction technique.
Have I done it correctly? If not, where did I make errors and how should I do it?
Thank you in advance!
Sorry for the bad image quality; I'm bad at taking pictures.
On
Just a little uncomfortable with step 6 of part 2, i.e., using three steps (1,4,5) to assert a contradiction - seems like too much intuition for my liking. If possible, would like to limit myself to just the { φ ∧ ¬φ } form of contradiction. Minimal intuition - an object cannot be both itself and not itself at the same time.
Is the following possible instead? {n}, where n is a natural number, denotes underlying dependencies.
{1} 1. ¬P ∨ ¬Q premise
{2} 2. P ∧ Q assumption for proof by contradiction
{2} 3. P 2 ∧E
{2} 4. Q 2 ∧E
{5} 5. ¬P assumption, 1st disjunct of ¬P ∨ ¬Q
{2,5} 6. P ∧ ¬P 3,5 ∧I
{7} 7. ¬Q assumption, 2nd disjunct of ¬P ∨ ¬Q
{2,7} 8. Q ∧ ¬Q 4,7 ∧I
{1,2} 9. (P ∧ ¬P) ∨ (Q ∧ ¬Q) 1,5,6,7,8 Discharge 5,7 i.e. ⊥
{1} 10. ¬(P ∧ Q) 2,9 proof by contradiction, discharge 2
Since step 10 is an outcome of the premise alone, therefore (¬P ∨ ¬Q) ⊢ ¬(P ∧ Q).
11 Feb: Just an additional comment after some thought. Citing steps 1 (¬P ∨ ¬Q), 4(P) and 6(Q) to justify a contradiction is implicitly claiming that (¬P ∨ ¬Q) is in contradiction with (P ∧ Q) (i.e. conjunction of steps 4 and 6). But this contradiction is the very thing we're trying to prove. That's why I wasn't comfortable previously. Glad for comments/correction if any. Thanks!
In the second part lines 2 and 7 are redundant, instead assume line 3.
Edit: As noticed by @GitGud, in the first part, line 6 must be $\neg\neg p \ (\neg I)$. So you need to add another line to get $p$ by $(\neg\neg E)$. Similar correction for line 10.