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

19 Views Asked by At

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.

1

There are 1 best solutions below

0
On

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.