# 7.2: Argument by Cases

Once we see how much work derived rules can save us, we will want others. Indeed, many derivations which are prohibitively complicated if we use only primitive rules become quite manageable when we can use derived rules also. Here is another example of a derived rule:

Argument by Cases

| XvY
| X⊃Z
| Y⊃Z
| Z AC

Here is a proof of this derived rule:

.
.
.
1 | XvY (Input for the derived rule)
2 | X⊃Z (Input for the derived rule)
3 | Y⊃Z (Input for the derived rule)
4 | | ~Z A
5 | | | X A
6 | | | X⊃Z 2, R
7 | | | Z 5, 6, ⊃E
8 | | 3 | ~Z 4, R
9 | | ~X 5-8, ~I
10 | | XvY 1, R
11 | | Y 9, 10, vE
12 | | Y⊃Z 3, R
13 | 2 | Z 11, 12, ⊃E
14 | ~~Z 4-13, ~I
15 3| Z 14, ~E

Again, the point of this proof is this: Suppose you have just used the derived rule Argument by Cases. Then, if you really want to, you can go back and rewrite your derivation using only the primitive rules. This proof shows you how to do it. Whatever sentence you have used for X and whatever sentence you have used for Y, just substitute the above in your derivation, after having written in your sentences for X and Y. Of course, you will have to redo the line numberings to fit with those in your full derivation. 1 have put in line numbers above to help in talking about the proof.

(A small point to notice in this example: In line 14 I have appealed to subderivation 2, lines 4-13, to use negation introduction. But where in the subderivation did I conclude both a sentence and its negation? The point is that the assumption can count as one of these sentences. Why? Because any sentence can be derived from itself, by reiteration. Thus, in derivation 2, I can fairly enough count both ~Z and Z as following from ~Z.)

Here is another derived rule, which we will also call Argument by Cases because it differs only superficially from the last:

Argument by Cases (second form)

| XvY
| .
| .
| .
| | X A
| | .
| | .
| | .
| | Z
| .
| .
| .
| | Y A
| | .
| | .
| | .
| | Z

| Z AC

In words, if in a derivation you already have a sentence of the form XvY, a subderivation from X as assumption to Z as conclusion, and a second subderivation from Y as assumption to Z as conclusion, you are licensed to write Z as a conclusion anywhere below.

This second form of argument by cases is actually the form you will use most often.
The proof of the second form of argument by cases goes like this:

1 | XvY (Input for the derived rule)
| .
| .
| | X A
| | . (Input for the derived rule)
| | .
| | Z
2 | X⊃Z ⊃I
| .
| .
| | Y A
| | . (Input for the derived rule)
| | .
| | Z
3 | Y⊃Z ⊃I
| Z From lines 1, 2, 3 and the first form of Argument by Cases

Note that in this proof I have used a previously proved derived rule (the first form of argument by cases) in proving a new derived rule (the second form of argument by cases). Why should I be allowed to do that, given that a proof of a derived rule must show that one can dispense with the derived rule and use primitive rules instead? Can you see the answer to this question?

Suppose we want to rewrite a derivation which uses a new derived rule so that, after rewriting, no derived rules are used. First rewrite the derivation dispensing with the new derived rule, following the proof of the new derived rule. The resulting derivation may use previously proved derived rules. But now we can rewrite some more, using the proofs of the previously proved derived rules to get rid of them. We continue in this way until we arrive at a (probably horrendously long) derivation which uses only primitive rules.

Argument by cases is an especially important derived rule, so much so that in a moment I'm going to give you a batch of exercises designed exclusively to develop your skill in applying it. Its importance stems not only from the fact that it can greatly shorten your work. It also represents a common and important form of reasoning which gets used both in everyday life and in mathematics. In fact, many texts use argument by cases as a primitive rule, where I use what I have called disjunction elimination. In fact, given the other rules, what I have called argument by cases and disjunction elimination are equivalent. I prefer to start with my form of disjunction elimination because I think that students learn it more easily than argument by cases. But now that your skills have developed, you should learn to use both rules effectively.

Exercise

7-1. Use argument by cases as well as any of the primitive rules to construct derivations to establish the validity of the following arguments:

a) AvB b) Av(BvC) c) (AvB)&(B>C) d) (A&B)v(A&C)
BvA (AvB)vC AvC A&(BvC)

e) A&(BvC) f) Av(B&C) g) (AvB)&(AvC) h) KvL
(A&B)v(A&C) (AvB)&(AvC) Av(B&C) K≡L
K&L

i) (D⊃G)v(D⊃I) j) ~CvK k) ~HvM l) (S&J)v(~S&~J)
D⊃(GvI) A⊃D ~M⊃~C S≡J
(AvC)⊃(KvD) (HvC)⊃M

m) K⊃(Fvc)
J⊃(CvD)
~C
~(FvD)⊃~(KvJ)

7-2. Show that in the presence of the other primitive rules, vE is equivalent to AC. (Half of this problem is already done in the text. Figure out which half and then do the other half I) , Biconditional Introduction Biconditional Elimination and Disjunction Elimination Denying the Consequent Reductio Ad Absurdum Vraditionally called "Modus Tolens")