Working with Conditional Introduction

        In the introduction we provided some examples of -›I. Here we will study it in more detail. Recall that it is a mode of proof. When you apply it you are appealing to the subderivation as a whole, not simply to the two lines you highlight in the builder. The builder indicates this by placing a '-' rather than a ',' between the numbers of the two lines that you highlighted. Once an assumption has been discharged the lines in the subderivation are inaccessible. Generally, though not invariably, you will set up a conditional introduction when the sentence you are trying to derive is a conditional or when you have need of a conditional as a step along the way toward your conclusion. Here is an example of the former where we show R |- P -› R:

1. 2. 3. 4.

R P R P -› R

Premise * A * 1, Reit 2-3 -›I

You should now:

Go to Lesson 1

        Here is another slightly more complex example where we show that (R v S) & ~R |- P -› (S v Q):

1. 2. 3. 4. 5. 6. 7. 8.

(R v S) & ~R R v S ~R S      P      S      S v Q P -› (S v Q)

Premise 1 &E 1 &E 2,3 DS * A * 1, Reit * 6 vI 4-6 -›I

Note here that we could have constructed a slightly shorter derivation if we had postponed the application of DS until after we had made the assumption. But you will likely find, when the derivations are more complex than the ones we have so far looked at, that applying the rules initially will help. You will see exactly what you have to work with and this may enable you to see a way to proceed in the derivation that you might otherwise have overlooked.

You should now:

Go to Lesson 2

Here is an example of the latter, that is, a derivation in which we use an -›I in order to facilitate obtaining something else:

1. 2.

(P -› Q) -› R Q

Premise Premise

Let us suppose that we are trying to show that (P -› Q) -› R, Q I- R. Note that if we were to have P -› Q as an accessible line then we could obtain the conclusion we want via an application of ›E. Since that is a conditional we can try to obtain it by an application of conditional introduction. Here is the completed derivation:

1. 2. 3. 4. 5. 6.

(P -› Q) -› R Q      P      Q P -› Q R

Premise Premise * A * 2 Reit 3-4 -›I 1, 5 -›E

You should now:

Go to Lesson 3

        Here is another slightly more complex example, but one that utilizes much the same tactic. You wish to show that ~P v (R & S), P, ((A & B) -› S) -› C I- C v D. Notice that the disjunct D does not occur in the premises. So you might get to the disjunction by applying vI to C. Looking at the premises you may note that C is the consequent of a conditional. So you might figure that you should try to use a subderivation to obtain that conditional.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

~P v (R & S) P ((A & B) -› S) -› C ~~P R & S S      A & B      S (A & B) -› S C C v D

Premise Premise Premise 2 ~~I 1, 4 DS 5 &E * A * 6 Reit 5-6 -›I 3,9 -›E 10 vI

While this derivation is longer it is not significantly harder provided you have mastered the other rules of the system. Notice that for the most part this derivation involves applying these other rules prior to making the assumption. The author generally prefers to do things this way since, as was noted above, we see what we have to work with. But you could just as well have made the assumption right after the premises. That derivation would look like this:

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

~P v (R & S) P ((A & B) -› S) -› C      A & B       ~~P      R & S      S (A & B) -› S C C v D

Premise Premise Premise * A * 2 ~~I * 1, 5 DS *. &E 4 - 7 -›I 3, 8 ›E 9 vI

Both of these are correct derivations. You should now:

Go to Lesson 4

Before continuing to the next tutorial or the exercises let us try one more derivation. You should now:

Go to Lesson 5

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