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I'm seeing math logic and I have a question.

Let $p$ be a proposition. Let's suppose I have $\lnot p$.

By disjunction rule, this implies $\lnot p \vee q$, where $q$ is any proposition.

This is equivalent (looking at the truth tables) to $p \implies q$

Does this mean that we only have to prove $\lnot p$ in order to prove $p \implies q\;$?

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    Let $p \equiv ( 2b^2=a^2 )$ and $q \equiv (a,b\notin \mathbb N$) now how does $\neg p$ proves $p \implies q$?2011-06-12
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    See my post below; the point is the statements p->q and ~p\/q are logically equivalent, i.e., they're either simultaneously true or simultaneously false, butthe logical equivalence does not extend to the existence of a proof of one from the other. So if you are given ~p, you can conclude ~p\/q. Then, any truth-value assignments to p,q will give the same truth values to the two sentences. In other types of implication, you may require that one sentence follows from the other to conclude p->q. Material implication may be part of the price to pay for the bakcboxing that goes on in sentence logic.2011-06-12
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    @ gary Logical equivalence between "p" and "q" DOES extend to the proof of "p" from "q" and conversely for classical propositional calculus and classical predicate calculus. The respective *completeness theorems* imply precisely that. @ Arjang Neither of what you meant by "p" nor "q" in your comment qualify as propositions here, so I don't see how what you wrote comes as relevant.2011-06-12
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    @gary Actually, that should go that the respective completeness theorems, *along with* the converse of the deduction theorem we have that from the logical equivalence of "p and "q", we have the existence of a proof of one from the other. For if |=Epq, the by the completeness theorem we have |-Epq. By equivalence elimination we can obtain |-Cpq and |-Cqp. Then by the converse of the deduction theorem we obtain p|-q, and q|-p. So "p" yields "q" and "q" yields "p", given that p and q come as semantically equivalent in this context.2011-06-13
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    @Doug: I took these two statements to be "blackboxed" sentences, and I did not consider any predicates. AFAIK, in sentence logic, the inner-structure of the sentences do not matter; as sentences ~p\/q is equivalent to p->q , independently of the meaning assigned to either--and this leads to what some would consider flaws. This is what I meant; that--when p,q are sentences-- the content of the sentences ~p,q is not needed in order to conclude p->q , while in predicate logic this is not the case.2011-06-13
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    You can click accept on the appropriate answer. The tick mark is next to the vote arrows.2011-06-13

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