Jan 2, 2017 · negation only applies to propositions. (p v q) is a proposition, call it r, so read ~ (p v q) as "it is not the case that the proposition r is true". p and q are also propositions, so e.g. ~p …

I would be grateful if someone could derive, by showing the proofs that: ~(P&Q) derives to ~Pv~Q. The same derivation would be appreciated for |- [(P>Q)>P]>P

Jan 3, 2020 · I understand the reasoning for why pVq implies ~p->q, but not the converse. What reasons are there to believe ~p->~q implies pV~q, other than to make the whole material …

Here is a way to format the proof so that it might make it easier to see the structure. I basically followed your lead. The question appears to be this: Am I allowed to use the conclusion at 14 …

Here's another of Tomassi's exercises I can't solve (Logic, page 106): ~ (~P & ~Q) : P ∨ Q I have to use natural deduction and the only rules I know are: assumptions, modus ponendo …

In the Stanford Truth Table Generator I used the following input strings to generate the three truth tables you presented as examples. ( (P->S)&& (PvQ)&& (Q->R))-> (SvR) ( (R&&S)&&S)-> (P) …

Nov 20, 2016 · In step two, I am not sure how you can use →-intro for that. For →-intro you need to point to a subproof with premise p and conclusion q, do you not? In my fitch program it does …

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 …

Aug 24, 2022 · Isn't that already the proof? As PvQ is true and as P is not true you know that Q must be true and as from Q follows not R you get not R from which follows not S qed. So the …

Sep 1, 2017 · Here's another Tomassi's problem I can't solve (Logic, Exercise 3.9.1.17, page 106): P ∨ Q : ~ (~P & ~Q) I have to use natural deduction and the only rules I know are: …