Earlier I noted the following crucial constraint on the use of CP: in any application of CP, the dependency-number of the antecedent must always be included among the dependencies of both lines used for CP. Otherwise, the consequent has not been derived from the antecedent. The proofs we have considered so far have posed no problems in this respect. Looking back, the common dependency-number (of the antecedent) is obvious. But that may not always be the case. For example, suppose I want to prove that:
Q P → Q. Given that the conclusion to be proved is a conditional, the strategy to crack the proof is CP. So, assume the antecedent and derive the consequent. But note carefully what results:
{1} 1. Q Premise
{2} 2. P Assumption for CP
3. P → Q ???
First, we introduce the premise, Q. Next, we assume the antecedent, P, for CP. We now want to apply CP. But we cannot: there is no number common to the sets of dependency-numbers for lines 1 and 2. So, the formulas on lines 1 and 2 have nothing whatever to do with one another. To apply CP, the formulas must work together such that we derive the consequent from the antecedent. Only in that way will we ever get the desired common dependency-number that will allow us to apply CP. In practice, this stumbling block is easily overcome. We can always get the formulas to work together by &I-ing them together and then &E-ing them apart again, i.e. we exploit the rules for ‘&’ to get the dependency-numbers just as we want them, like so:
Q P → Q
{1} 1. Q Premise
{2} 2. P Assumption for CP
{1,2} 3. P & Q 1,2 &I
{1,2} 4. Q 3 &E
{1} 5. P → Q 2,4 CP
Note that the formulas on lines 1 and 4 are the same, namely Q. But note also that each occurrence of the sentence-letter Q has different dependency-numbers. The set of dependencies belonging to line 4 is larger than that of line 1 because the formula on line 4 is the result of conjoining Q with P and then taking Q back out of that conjunction. In virtue of those moves, the rules do now permit us to apply CP to lines 2 and 4 just because there is a common number, namely, 2. That number is precisely the dependency-number of P and P is precisely the antecedent.
It is tempting to describe this useful strategy as ‘cheating with CP for exam purposes’. However, according to certain logicians, known as
‘relevance’ logicians, if one cheats in order to get the right answer by illicit means then this is worse than cheating, for in their view it is just to get the wrong answer.
That is not to say that relevance logicians object to the rules of &I and
&E. These rules are not considered problematic in themselves. Rather, the objection is that exploiting the rules in that way legitimates CP on the result. Again, however, this is controversial territory and, for our purposes, this strategy for CP is a perfectly legitimate one. Indeed, it has the solemn and impressive title of augmentation, i.e. Q may be augmented by P using the & rules. Moreover, the legitimacy of augmentation as a strategy is underpinned by the definition of validity which we stated in Chapter 1. Remember: the truth of the premises is sufficient to guarantee the truth of the conclusion. But if the conclusion
follows from some true premises and we simply add a few more true premises the conclusion must still follow. More generally:
If a given formula follows from a particular set of formulas then the same formula follows from any augmentation of that set of formulas.
This property of PL is known as monotonicity: if a given formula follows from some set of formulas then any addition of further formulas to that set results in a set from which the original formula will still follow. Fortunately for those facing exams in PL, PL is a monotonic logic.
IX
Theorems
CP is the first rule of inference we have encountered which allows us to discharge assumptions. But it will not be the last. Discharge rules are especially useful for proving sequents which represent the traditional laws of logic or logical truths in PL. Why should this be so? Remember that CP is a discharge rule which allows us to discharge assumptions, and that applications of that rule may be iterated. Suppose we only have assumptions to worry about in a given case. Further suppose that we can and do keep iterating CP until there are no assumptions left for the conclusion to depend upon. What are we to say of a conclusion derived in this way? Does it follow from nothing? After all, sequents representing such formulas will have nothing on the left-hand side of the turnstile, i.e. no premises will be involved. So, they will consist just of a turnstile followed by a formula of PL. Certainly, these formulas require no premises, but that does not mean that they follow from nothing. Rather, the mere form of such a formula is always sufficient to guarantee its truth.
Those sequents of PL whose set of premises is the empty set, i.e. which have no premises, are known as theorems of PL. The theorems of PL include the traditional laws of logic. And this is unsurprising. Theorems are logical truths. Recall the definition of validity: an argument is valid if and only if it is impossible for the conclusion to be false when the premises are true. But logical truths are never false. No matter how we interpret the sentence-letters involved in any logical truth it still comes out true.
For example, consider the law of identity: P → P. There is no possible interpretation under which the formula P → P is false. If it’s raining then it’s raining. If the sun is shining then the sun is shining. And so on, ad nauseam.
The very form of this formula ensures that no matter how we translate or interpret its constituent sentence-letters it will always be true. Hence, the
law of identity is a logical truth. It is also a theorem of PL and because the formula is a conditional we prove it by CP. In the following proof of the law of identity note how a well-formed formula in effect takes itself as assumption. This is a perfectly respectable application of CP and one which is eminently useful at times:
P → P
{1} 1. P Assumption for CP
— 2. P → P 1,1 CP
Keep this strategy in mind when you attempt Exercise 2.6 (particularly when you try to prove the first theorem).
EXERCISE 2.6
1. Prove that the following sequents are theorems of PL:
1. :((P → P) → Q) → Q (5)
2. :(P → Q) → ((Q → R) → (P → R)) (8) 3. :(Q → R) → ((P → Q) → (P → R)) (8)
4. :P → (Q → (P & Q)) (5)