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".

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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 implication system neat?

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The argument schema is not valid; check it with both P and Q false.. Thus, we cannot prove that (¬P & ¬Q) : P v Q.

Sep 1, 2017 · @DiegoRuizHaro You have reductio ad absurdum (RAA) which is similar. I don't have that rule in the proof checker that I am using, but I imagine it goes the same way except instead of introducing the contradiction and then discharging the assumption with an indirect proof, you may have to reach something like P & ~P and then discharge the assumption with an RAA rule of some sort.

(((pvq)->r)&&(q))->(r) Each truth table generator will have its own input syntax so you will have to be careful to follow that. Regardless, take each premise and surround it with parentheses and then "and" them together.

Aug 24, 2022 · 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 problem isn't were to start but how to write that in this weird system right?

Apr 19, 2016 · I think that's more of a feature of modal logic, not the operator itself. For instance you could say (PvQ), ⋄(PvQ), or (PvQ). The function of "v" won't change, but the modal operator changes the truth conditions of the expression within its scope. I'm not too sure if what I said is true though haha. –

Mar 9, 2016 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.