Tutorial 25 ... | SoftOption ®

Logical System

12/26/13

You now have the tools to appraise arguments to the level of detail offered by predicate logic.

Let us run through how these might be used with a long and difficult example.

Consider the argument

First we symbolize it

Then we should have a quick look at the semantics. We start by producing an interpretation in which the conclusion is false, say

Now the first four premises are true in this, but the fifth one is not, we need for (∃x)(Cx∧∼Vx) to be true (ie for there to be an x which is C but not V). We try

Now, the conclusion is true, and so too is premise 5, but premises 2 and 3, (∀x)(Cx⊃(Fx∨Ix∨Px)) and (∀x)(Cx⊃Kx), are not true.

We need to make our a have the property K to get (∀x)(Cx⊃Kx) to be true, and it must also have one of the properties F, I, or P to get premise 2 to be true. This is all starting to get a little sticky.

At this point we should wonder whether the argument might be valid, and whether we can produce a derivation of it. The derivation starts

We have used Tactics, and it seemed likely that the last line has come from Existential Instantion (which we have put in). (There is a little hint here. If you have a whole lot of universally quantified formulas, and a single existentially quantified one, try working with instantiating the existential quantifier first (that gives you a suitable term to plug into the universally quantified formulas). Line 5 offers an invitation.)

Tactics, EG. Now our problem is to get from line 6 Cx∧∼Vx to line 8 Cx∧Ix. The Cxs will take care of themselves, so the problem is to get from∼ Vx to Ix. The only place that I appears is on the right hand side of 2. So we should take the Cx, which we need

and then the right hand side of 2 (and 3 while we are working on Cx)

Now we are still trying to get Ix from a variety of things, but most obviously line 9 Fx∨(Ix∨Px) which suggests an ∨E

At this point we are not quite sure what to do. Sometimes a way forward is by finding a contradiction and using reductio. We have ∼Vx at line 6 and the right hand side of 1 is Vx, so this might be a way to go.

Then

We finish the reductio and are looking at a second ∨E

now all that is left is getting from Px to Ix

This is a bit of a puzzle (so again reductio might be in order). Another little hint or observation here is that we have not used line 4 at all (you don't have to use all the premises, but they are a resource for you). Line 4 is ∼(∃x)(Kx∧Px) which cannot be used directly on its own. But it has a negation as its main connective which means it may play a role in reductio (especially when we realize that we have both Kx and Px at lines 11 and 24).

We do the steps, including EG, to produce the positive side of the contradiction

We set out on the reductio via the double negation.

Laying it out.

Finally, completing the proof

The example argument is valid. The derivation that proves that it is valid is long and difficult (it might have been easier if rewrite rules had been used). Also, derivations are not unique. There are can be many different proofs of a theorem (or that an argument is valid). There may even be an 8 line proof of the above!

Exercise to accompany Predicate Tutorial 15.

Exercise 1(of 1)

Most experienced logicians are pretty familiar with the following theorems. Although the theorems are stated in terms of Fx and Gx etc. most of them hold of most formulas. Derive them. [You may use rewrite rules for these, if you wish.]

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Interpretations Widget

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