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If one wants to proof $P\vee Q$, is it sufficient to proof $\lnot P \rightarrow Q$? Because it makes intuitively more sense to me that $P\vee Q$ would be logically equivalent with $(\lnot P \rightarrow Q) \wedge (\lnot Q \rightarrow P)$.

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It is in fact equivalent to $\lnot P\rightarrow Q$, which is also equivalent to $\lnot Q\rightarrow P$. So yes, it is indeed sufficient. However, in a formal logic system you should probably prove this using whatever axioms/rules of deduction you have.

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  • $\begingroup$ How stupid of me not to think of that. Thanks. $\endgroup$ – Michael Angelo Jan 27 '15 at 10:14

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