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33.4 Annoying Pattern ~AxP(x)

This is the last critical pattern you need to know!

In the previous section we saw that ~ExP(x) is a lot like the 5-step plan for ~(PvQ).

~ExP(x) is analogous to ~(PvQ).
~AxP(x) is analogous to ~(P&Q).

The key to understanding ~AxP(x) is to realize that it is analogous to ~(P&Q).

Recall that ~(P&Q) was an annoying pattern, because we could not apply the 5-step plan to it directly. That’s because we can’t do &Intro with just one conjunct, after assuming P, the way we can do vIntro.

Let’s make sure you can still do the DeMorgan’s proof  from ~(P&Q), so that you know what we are talking about.

The key idea to this proof is to realize that we have to start with a reductio.

That gives us negation around a disjunction, which is a pattern we already know how to deal with.

So even though we cannot apply the 5-step plan to ~(P&Q) directly, we still use the 5-step plan indirectly, by applying it to the reductio assumption on line 2, ~(~Pv~Q).

We are in a similar situation with ~AxP(x). We know by DeMorgan’s for quantifiers that that entails Ex~P(x).

We cannot deal with ~AxP(x) directly.

But we cannot deal with ~AxP(x) directly. We have to do a reductio, by assuming ~Ex~P(x).

Notice that that gives us a negation around an existential, and you do know how to deal with that. You just learned it in the previous section!

You should try to figure it out yourself, so grab a piece of paper now and see if you can do this proof now.

1. ~AxP(x)
Thus,
2. Ex~P(x)

Afterward, keep reading below.

***Continuing***

We need a reductio, so we start by assuming ~Ex~P(x) on line 2.

We know that on the second-to-last line of the proof, just before the conclusion, we’ll end that subproof with #, since that is how we finish a reductio.

How are we going to get #?

But how are we going to get that #?

You might think we get it by building ~Ex~P(x) from the inside, by proving Ex~P(x). But that is not quite right.

Look back at the proof starting with ~(P&Q). We got the final contradiction by building P&Q. That is, by building the original premise from the inside.

In the current proof the premise is ~AxP(x). So what we need to prove is AxP(x).

What is nice about that is it is a wide-scope universal, which tells us what to do next: plan out a AIntro proof.

We need to assume @n on a new subproof, and we need to end  that subproof with P(n).

If you’ve followed us this far, you might be able to complete the proof yourself, so let’s try it!

Dealing with negation around a universal is no easy feat, so if you could figure out that proof, great work!

Resolve this proof regularly until you’ve mastered it!

If you didn’t get it yet, don’t worry. What you need to do is return to this section regularly, reviewing and retrying these ideas until you’ve got it.

It’s also helpful to practice similar but slightly different proofs. Here’s another one you can try.

If you are struggling with this proof, head to section 33.6 and try the related proof there first.