Choosing Logical Connectives 1. Too Few Connectives?: We have chosen to use only 5 logical connectives in our constructed language of logic, L1 (they are:,,,, and ). But, we might ask, are these enough? Consider: For 2-line truth tables, L1 only has ONE connective (namely, ). But there are actually FOUR possible connectives that operate on single atomic statements: P C1 (%P) C2 (@P) C3 ( P) C4 (#P) T T T F F F T F T F In other words, we could have chosen to include a connective in our language L1 (call it # ), which is such that, when P is true, #P is false, and when P is false, #P is false. Why didn t we? But, it gets worse: For 4-line truth tables, which operate on 2 sentence letters, we have chosen to use only 4 connectives (namely,,,, and ), but it turns out that there are actually SIXTEEN possible 2-letter connectives: P Q C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12 C13 C14 C15 C16 And there are actually 256 possible connectives which operate on 3 letter-sentences (the number of possible connectives will be 2 raised to the 2 n power (i.e., 2 2^n ), where n is the number of atomic statements being operated on), but L1 contains zero of them. Indeed, there are an infinite number of possible logical connectives that we could have chosen to include in L1, but we chose only FIVE! When we ask, Are five connectives enough?, what we are really asking is, Is L1 expressively complete? That means: Expressively Complete: A formal language, L, is expressively complete iff any wff containing any truth-functional connective can be translated into a wff that is logically equivalent to it, but which contains only connectives that are part of L. 1
In other words, for any of the 20 connectives we just saw truth tables for, L1 should be able to express a symbolized statements that operates on P and Q in those ways, to deliver exactly those truth tables. If L1 CANNOT do that, then there are some statements that L1 simply cannot express and that is bad It would mean that L1 is incomplete. But, let s look at the 16 possible 2-letter operators again: P Q C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12 C13 C14 C15 C16 Are any of these 16 connectives familiar? Four of them should be. They are: C3 = C7 = C9 = C15 = Additionally, four of the connectives above are the contradictories of these four. For instance, compare C8 with C9. They are contradictory. That is, their truth values are the opposite on every line of the truth table. So, since C9 represents P Q, C8 represents (P Q). Take a look and try to find the contradictories of the other three connectives. C3 s contradictory is C14. C7 s contradictory is C10. And C15 s contradictory is C2. So, now we have: P Q C1 (P Q) P Q C4 C5 C6 P Q (P Q) P Q (P Q) C11 C12 C13 (P Q) P Q C16 Let s look at some of the others. Start with C5. What sort of formula involving P and Q would ONLY come out false when Q is true and P is false? That should look familiar. It s Q P of course! So, C5 represents the truth function of Q P, and therefore, it s contradictory, C12, represents (Q P). Now take a look at C13. It simply has the same truth function as P. And C11 is just Q. So, in turn, their contradictories (C4 and C6) are just P and Q. 2
P Q C1 (P Q) P Q P Q P Q P Q (P Q) P Q (P Q) Q (Q P) P (P Q) P Q C16 All that are left are C1 and C16. Notice that C1 is just a tautologous statement and C16 is just a self-contradictory statement. (That is C1 always comes out true, and C16 always comes out false). So, for C1, we need nothing more than P P, while for C16 we just need P P. We re done! P Q P P (P Q) P Q P Q P Q P Q (P Q) P Q (P Q) Q (Q P) P (P Q) P Q P P Conclusion: Assuming we could continue in this way, representing all 256 of the 3-letter operators, and the infinite number of 4-letter, 5-letter, etc. operators (and, the short story is, we COULD do this), it turns out, L1 is expressively complete. Think about that: With just 5 little operators, L1 is able to express an infinite number of English sentences! 2. Too Many Connectives?: Really, though, we could make do with LESS THAN five operators. One of the five in particular should stand out as unnecessary. Which one? The bi-conditional! In fact, when we defined the bi-conditional, we said that it was really just two conditionals. In other words: P Q (P Q) (Q P) So, the is really unnecessary. It can just be expressed using and, since the two statements above are logically equivalent. So, we could have chosen only 4 operators instead of 5. Are there any other unnecessary operators? Well, it turns out that can be expressed in terms of and, such that: P Q (P Q) 3
For instance, if I say, If Fluffy is a cat, then Fluffy is a mammal, this really means the same thing as, So, it must not be the case that Fluffy is a cat AND NOT a mammal. Note: If you question this, then draw up a truth table for (C M) and see that it is the same as the truth table for C M. So, we don t need either. L1 would still be expressively complete even if it just included, and. We could have chosen to express and as follows: P Q (P Q) P Q (P Q) (Q P) But, wait. The isn t necessary either! It s reducible to and, since: P Q ( P Q) For instance, if I say that Timmy is either five or six, this really just means that It must NOT be the case that Timmy is not five and not six. Note: Again, if you question this, then draw up a truth table for ( F S) and see that it is the same as the truth table for F S. So, we don t need either! We ve done away with all but TWO operators! That is, L1 would still be expressively complete if the only two operators were and. The Sheffer Stroke: But, it gets crazier. Henry Sheffer (in 1913) pointed out that a language could be expressively complete with just ONE logical connective. It s called The Sheffer Stroke. We represent it with a slash,, and its truth table corresponds to C2, above. So, it turns out that P Q is logically equivalent to (P Q) : P Q (P Q) P Q T T F F T F T T F T T T F F T T In other words, P Sheffer-Stroke Q (or P Q ) just means Not both P and Q. For instance, we might symbolize Albert and Bertha weren t both at the party as A B. 4
Note: Notice that, whereas we reduced all 5 logical connectives to 2 connectives, and, Sheffer reduces all 5 connectives to ONE connective simply by introducing a NEW connective that is logically equivalent to (P Q), which contains both and. Sheffer showed that we can reduce all 5 of the operators to the Sheffer Stroke: P P P P Q (P Q) (P Q) P Q (P P) (Q Q) P Q P (Q Q) P Q {[P (Q Q)] [Q (P P)]} {[P (Q Q)] [Q (P P)]} Duuuuude, whooaah.. Conclusion: So, if there are an infinite number of possible logical connectives, but every sentence of L1 can be expressed with only ONE connective, why did we choose EXACTLY FIVE connectives no more, no less? The answer is that, we want to strike a balance between efficiency and parsimony. A parsimonious system uses as few connectives as possible. But, TOO FEW connectives, and the system becomes too burdensome to be efficient. (To see this, just look at how crazy the simple claim P Q becomes if we re only using the Sheffer Stroke!) We can express sentences with simpler, more efficient symbolizations by accepting MORE connectives but if TOO MANY connectives, and the system is no longer very parsimonious. So, we have to strike a balance, and arbitrarily pick some number of connectives that is large enough to not result in sentences that are too crazy and complicated, but small enough so that we are not working with an overly-long list of connectives. Logicians typically settle on the 5 connectives we have been using, no more, no less. 5