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I have no special skills of doing this. Can you introduce how to think of that ? I could take Xv!X as hip and then proof by parts x -> (X->Z)v(Z->X). But is the best way always to split disjunction ?

Maybe some kind of simple learning resources ? Thanks.

Timotei
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  • I assume "v" means $\lor$, and "!" means $\neg$ (negation). You can proceed as follows. Take $(X\vee\neg X)$ as hypothesis, infer $(\neg X\vee X)$ and split into cases: if $\neg X$, then $X\to Z$ (why?); if $X$, then $Z\to X$; so $(X\to Z)\vee ($Z\to X$)$; now discharge the hypothesis: voila, your theorem. // By the way, what you've exhibited in the title is not an "equivalence" it's an implication. – BrianO Nov 11 '15 at 10:22
  • I don't know the rules of inference or the axioms that we can use. So, I simply can't give you much with regard to deriving this in propositional logic. – Doug Spoonwood Nov 11 '15 at 15:50

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