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I'm teaching myself basic logic. I've answered all the other problems in the section, most rather easily, but this one.

$$(B\to{\lnot}C)\to D\quad {\therefore}B\to({\lnot}C{\to}D)$$

original problem image

(Note. The original version of this question read "$(B\to{\lnot}C)\to\lnot D$". It has been edited by @Blue to match the problem image, which may-or-may-not be what the asker intended.)

I feel as though I'm probably just starting with the wrong assumptions, although I've tried a few. Any small hints would be appreciated.

Blue
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Charles J. Bates
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  • I transcribed the problem from the image you gave, in doing so, I removed a "$\lnot$" from your version of the problem statement. Please double-check which version is the one you intended. (If you were working with the wrong one, then it's not much of a surprise that you got stuck. :) ... In general, it's a good idea to include the work you've done. This helps answerers tailor their responses to help you without duplicating your effort. Plus, it should help resolve questions about which version of the problem is correct. – Blue Oct 17 '19 at 03:21

1 Answers1

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The goal is a conditional, so do a conditional proof: assume $B$, and try to get to $\neg C \to D$

Ok, but that new goal is itself a conditional, so do a second conditional proof inside the first one: assume $\neg C$, and try to got to $D$

Now, given the premise, it is clear that you can get to $D$ if only you can prove $B \to \neg C$. So, the new goal is to prove $B \to \neg C$ .... which is yet another conditional, so do a third conditional proof inside the second: assume $B$ and try to get to $\neg C$. But, you have $\neg C$ as the assumption of the second subproof, so the you can just reiterate that, completing the third conditional proof. And now you can wrap everything up as planned.

Bram28
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