Revised 4/14/99

Sometimes it is convenient to use a technique called a "conditional proof" (CP). This device allows one to assume a proposition, then infer something from it (and any other available propositions). The assumed proposition and the resulting proposition are then linked together in a conditional statement. Perhaps the resultant conditional could have been arrived at in some other way, but a CP is often more convenient. Also, the method can be extended and adapted into a very powerful technique which will be presented in the next lesson--indirect proof.

Here is the ordinary reasoning process that is captured by CP. Say, you are asked to go somewhere--a concert in Atlanta. You then think, "Hmmm, if I go to the concert then I will get in very late or even the next day. But I have to be up early the next day to help Jim move. So if I go to the concert I won't be able to get up early to help Jim move like I promised. I don't think I'll go to the concert." You did not assert that you would go to the concert. Rather you thought, "If I go to the concert, then I will not be able to get up early to help Jim move." You considered a conditional statement. This hypothetical consideration then led you to conclude that you should not go to the concert.

CP allows one to do this sort of reasoning in a derivation. But there are certain restrictions. One can make any assumption that one wants (similar to the rule for Addition). But one must make this assumption within a limited context and clearly indicate what one is doing. Also the assumption leads to a conditional statement and not to any other kind of proposition.

Here is how it works. Let's say we have the following argument:

If I go to the concert, then either I get in late or I get in the next day and I won't be able to help Jim move.
I have to be up early the next day to help Jim move.
So I should not go to the concert.

Symbolized it looks like this:

C É [(L Ú N) & ~H]
H     
~C

Now let's work out the derivation, using a conditional proof:

1. C É [(L Ú N) & ~H]

2. H      /\ ~C

First we assume a C. By doing this we will be able to work towards the conditional that we will need: C É ~H. Then we will be able to easily get the ~C.

3.            C                                   Assume

This move is the first step in a conditional proof. It is indicated by setting it over to the right. Also we designate it as an assumption.

4.           (L Ú N) & ~H      1,3      MP

5.          ~H & (L Ú N)        4       Comm

6.          ~H                         5       Simpl

Nothing new here. Once we assume the C, we are able to make use of the preceding lines and the rules of inference and replacement to get a ~H. Now comes the other new move. We form a conditional on the basis of what we have done in lines 3 through 6. But the next line is not a part of the conditional proof; it is the result. So we put it over to the left.

7. C É ~H                       3-6        CP

NOTE: Every assumption must eventually result in a conditional proof (CP). (I say "eventually" because it is permissable to do a CP within a CP.) One cannot return to the main derivation until one has closed off the lines that begin with an assumption by asserting a conditional statement that follows from the conditional proof.

The rest of the derivation is unremarkable:

8. ~~H                              2       DN

9. ~C                              7,8      MT

Sometimes, as noted earlier, we can do a derivation without using CP, but CP is more convenient. Here is an example:

(A Ú B) É C      /\ A É C

I worked this one in eleven lines using Conditional (a couple of times), DeMorgan, Distribution, Commutation (two times) and Simplification. But notice how easy it is to do with CP:

1. (A Ú B) É C          /\ A É C

2.             A                         Assume

3.             A Ú B                 2         Add

4.             C                      3,1        MP

5. A É C                          2-4        CP

NOTE: By assuming A, then using Addition, I was able both to get the antecedent and the consequent for the conditional on line 5. If I had assumed A Ú B, then I would have been able to get C, the consequent, but not the A that I needed for the antecedent.

At any rate, I was able to get to the conclusion in 5 steps, using only Addition, Modus Ponens and a Conditional Proof. This was much simpler than the eight-line derivation that I did without CP. Plus the eight-step proof made use of many more rules, some of which are cumbersome, such as Distribution.

Exercise 14 provides practice opportunities.

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Copyright © 1999, Michael Eldridge