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It's really complex and I really need a simple way to do it. I just go blank when I see it. Please help...

Gotham Knight
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2 Answers2

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As a general strategy: if you ever need to prove a conditional, start a subproof where you assume the 'if' part, and try to get to the 'then' part as your last line of the subproof, after which you can close the subproof and infer the conditional using $\rightarrow \ Intro$

Here is a proof made in 'Fitch', a software program to make Fitch proofs:

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Bram28
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  • You made three hypotheses. If you have proof by contradiction {$\lnot$x ... {y, $\lnot$y}} $\vdash$ x, and the rules $\lnot$(x -> y) $\vdash$ x, $\lnot$(x -> y) $\vdash$ $\lnot$ y, it could get proved with a single hypothesis. – Doug Spoonwood Jul 23 '17 at 19:03
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The key is this: in order to prove a statement of the form $A \implies B$, begin by assuming $A$ and try to conclude that $B$ is true using any previous assumptions. Here's how we can do this proof:

[Want to prove: (p->(q->r))->(p->q)->(p->r)]
1. Assume: (p->(q->r))
    [Want to prove: (p->q)->(p->r)]
    2. Assume: (p->q)
        [Want to prove: p->r]
        3. Assume p
            [Want to prove: r]
            4. By 2, we can conclude that q is also true
            5. By 1, we can conclude that q->r.
            6. Because q and (q->r), conclude that r is true
            [By proving r with assumption p, we have proven that p->r]
        7. Thus, p->r
        [By proving (p->r) with assumption (p-> q), we have proven that (p->q)->(p->r)]
    8. Thus, (p->q)->(p->r)
    [we have now proven that (p->(q->r))->(p->q)->(p->r), as desired]
Ben Grossmann
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