@nemo66,

nemo66 wrote:(De Morgan's laws cannot be used).

It makes sense that DeMorgan cannot be used, as it is DeMorgan that is to be proved here!

I think for the first problem, there are missing parentheses and it should instead be (¬PvQ)→¬(P&¬Q), otherwise it would not follow.

We proceed here by using reductio ad absurdum.

Code:

1 (1) ¬PvQ assumption

2 (2) P&¬Q assumption (assume the negation of the conclusion)

3 (3) ¬P assumption

2 (4) P & elim

2, 3 (5) <contradiction> 3, 4 not-elimination

6 (6) Q assumption

2 (7) ¬Q 2, & elim

2,6(8) <contradiction> 6,7 not-elimination

1,2(9) <contradiction> 1,3,5,6, 8 v-elimination

1 (10) ¬(P&¬Q) 2, 9 not-introduction.

(11) (¬PvQ)→¬(P&¬Q) 1, 10 conditional proof □

In a reductio ad absurdum, making use of not-elim and not-intro rules is necessary. It cannot be done otherwise.

The second one ¬(P&Q)→¬Pv¬Q is as follows:

Code:

1 (1) ¬(P&Q) Assumption

2 (2) ¬(¬Pv¬Q) Assumption

3 (3) ¬P Assumption

3 (4) ¬Pv¬Q 3, v-intro

2, 3 (5) <contradiction> 2, 4 not-elim

2 (6) P 3, 5 not-intro (also double negation if allowed)

7 (7) ¬Q assumption

7 (8) ¬Pv¬Q 7, v-intro

2, 7 (9) <contradiction> 2, 8 not-elim

2 (10) Q 7, 9 not-intro (double negation if allowed)

2 (11) P&Q 6, 10 &-intro

1, 2 (12) <contradiction> 1, 11 not-elim

1 (13) ¬Pv¬Q 2, 11 not-intro ( and Double negation)

(14) ¬(P&Q)→¬Pv¬Q 1, 13 conditional proof □

If you allow me to advise you, then I advise you to try to work it out by yourself thoroughly before seeking help, otherwise you will not benefit fully from your studies. As very last resort seek help.