Subderivations

Conditional introduction? In class last session we looked at the conditional elimination rule. And in the textbook Teller took us through the rules for disjunction introduction (which says that if some P is true, then P ∨ Q will also be true, for any Q) and disjunction elimination (which says that if some P ∨ Q is true and ~P is true, then Q must be true). But what about conditional introduction?

Conditional introduction As Teller explains, to get the conditional introduction rule we first need the idea of a subderivation. How should we do this? Well, how would we normally argue if we wanted to persuade someone of the truth of some conditional, some ‘if X, then Y’?

Conditional introduction A good method would be to get them to assume that X is true, and then by a series of steps get them to admit that if X is true, then Y would have to be true as well. That way, even if we haven’t established the truth of X, we will have established the truth of ‘if X, then Y’.

Subderivation Subderivation is supposed to reflect this model of reasoning. Within our main argument, we start a sub-argument, in which we introduce the antecedent of the conditional as if it were a premise by assuming it. We then use whatever other premises we have to prove that the consequent would follow from the antecedent. Then we ‘discharge’ the original assumption and conclude the whole conditional.

Subderivation Let’s look at an example: 1 P ⊃ Q P 2 Q ⊃ R P 3 P Assumption (A) 4 P ⊃ Q 1, Reiteration (R) 5 Q ⊃ R 2, R 6 Q 3, 4, ⊃E 7 R 5, 6, ⊃E 8 P ⊃ R 3-7, Conditional Introduction (⊃I)

Subderivations and reiteration The subderivation is doing what we described earlier – we assume P, and then derive R, and from that we can conclude that P ⊃ R, and discharge the assumption that P. Lines four and five are just repeating two of our original premises in the subderivation. Since these are assumptions of our entire argument, we’re free to repeat (or reiterate) them wherever we like, even in subderivations.

The reiteration rule Here’s a formal rule that licenses reiteration: If a sentence occurs, either as a premise or as a conclusion in a derivation, that sentence may be copied (reiterated) in any of that derivation’s lower subderivations, or lower down in the same derivation.

Conditional Introduction OK, now we’re (finally!) in a position to introduce the conditional introduction rule. What we need is something graphical to reflect this: If you have a subderivation with assumption P and final conclusion Q, then P ⊃ Q may be entered below the subderivation as another conclusion of the outer derivation. (The subderivation may use any previous premise or conclusion of the outer derivation, entering these in accordance with the reiteration rule.)

Conditional Introduction . . . P Assumption (A) . . Q P ⊃ Q Conditional Introduction (⊃ I)

Assumptions vs Premises We call the thing introduced at the top of a subderivation an assumption rather than a premise. This marks an important distinction. The premises of an argument are there for good – we can take their truth for granted wherever we are in an argument. Assumptions we only suppose to be true temporarily. Eventually, when the subderivation ends, we discharge the assumption. From that point on, we cannot treat it like a premise – it drops out of the final picture.

Scope lines This is why scope lines matter. They tell us what we’re allowed to assume at different points in the argument. Assumptions only fall under the scope lines of subderivations. They are not assumed in the outer derivation. So once we’ve finished the subderivation and moved back to the outer derivation, we can no longer treat the assumption like a premise.

Scope lines All of this means that once you have been through a subderivation and then applied the conditional introduction rule and discharged your assumption, the conditional you end up will be true even if the assumption you used is false. All that the truth of the conditional rests on is the original premises of the outer derivation.

Exercises Provide a derivation that shows this argument to be valid: X ⊃ Y Y ⊃ Z Z ⊃ S X ⊃ S