So far we’ve learned the basic ideas for conditionals: what to do when you have a main connective conditional premise or conclusion.

If we can deal with ~(P->Q), we can deal with anything.

Next we turn to the annoying case: a negation around a conditional.

Remember from BOOL that the hardest patterns were when a negation symbol was wide-scope and blocked access to a disjunction or conjunction. We learned the 5-step plan to deal with those.

The contradiction trick (19.3) and the reiteration trick (19.4) are the PROP analogues of the 5-step plan.

Before we get to the details, it is essential to notice this idea: whenever you have a negation around some other information, you get at that information by building it from the inside.

This is what we mean.

Whenever you have a negation around some other information, you get at that information by building it from the inside for reductio.

We used the 5-step plan on this pattern: ~(PvQ). That plan is a reductio. We assume P, and then we build PvQ.

That is what we mean by building it from the inside. We build PvQ, the inside of the negation pattern, in order to extract information out of it by reductio. What we extract is ~P from the first disjunct and ~Q from the second.

Proving this DeMorgan’s was more annoying:

1 ~(P&Q)

Thus,

2 ~Pv~Q

Because the ~(P&Q) pattern requires an indirect approach. We had to do a reductio, and assume ~(~Pv~Q). But recall how that reductio worked: we eventually built P&Q in order to derive the contradiction.

So we again build the inside of ~(P&Q) in order to extract information from it by reductio.

That will be exactly how we deal with ~(P->Q).

What that means is we should be able to do this proof:

1 ~(P->Q)

Thus,

2 P&~Q

Even without knowing how to handle ~(P->Q), the main connective of the conclusion should tell you a plan.

We should be able to prove P alone, and prove ~Q alone, and then put them together with &Intro.

Knowing what you now do, about building things from the inside, let’s see if you can figure out how this goes.

If you couldn’t figure that out on your own, don’t worry. We’ll walk you through the thought process.

We cannot apply an Elim rule to ~(P->Q), and we have no other information to work with. So we have to do a reductio.

That’s how we know from the get-go that line 2 is ~P and line 7 is #.

But ~P is a literal. We can’t apply an Elim rule to it. And we can’t do a reductio of our new goal, #, because a reductio of # is never helpful.

Here’s the key idea we need to make progress: the way we are going to get # on line 7 is building ~(P->Q) from the inside. Namely, we build P->Q.

Here’s the key idea we need to make progress: the way we are going to get # on line 7 is building ~(P->Q) from the inside. That is, we build P->Q on line 6. That’s how we get # on line 7.

Lastly, we’ll point out that the reason why we call it the contradiction trick is not because of the # on line 7. The trick is knowing how to build P->Q. It’s the contradiction on line 4 that is the trick.

Now you can try another problem which uses the contradiction trick.

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