Truth-Functional Completeness

1. A set of truth-functional operators is said to be truth-functionally complete (or expressively adequate) just in case one can take any truth-function whatsoever, and construct a formula using only operators from that set, which represents that truth-function. In what follows, we will discuss how to establish the truth-functional completeness of various sets of truth-functional operators.

2. Let us suppose that we have an arbitrary n-place truth-function. Its truth table representation will have 2n rows, some true and some false. Here, for example, is the truth table representation of some 3- place truth function, which we shall call $:

Φ ψ χ $ T T T T T T F T T F T F T F F T F T T F F T F T F F T F F F F T

This truth-function $ is true on 5 rows (the first, second, fourth, sixth, and eighth), and false on the remaining 3 (the third, fifth, and seventh).

3. Now consider the following procedure: For every row on which this function is true, construct the conjunctive representation of that row – the extended conjunction consisting of all the atomic sentences that are true on that row and the negations of all the atomic sentences that are false on that row. In the example above, the conjunctive representations of the true rows are as follows (ignoring some extraneous parentheses):

Row 1: (P&Q&R) Row 2: (P&Q&~R) Row 4: (P&~Q&~R) Row 6: (~P&Q&~R) Row 8: (~P&~Q&~R)

And now think about the formula that is disjunction of all these extended conjunctions, a formula that basically is a disjunction of all the rows that are true, which in this case would be

[(Row 1) v (Row 2) v (Row 4) v (Row6) v (Row 8)]

Or,

[(P&Q&R) v (P&Q&~R) v (P&~Q&~R) v (P&~Q&~R) v (~P&Q&~R) v (~P&~Q&~R)]

4. It’s pretty easy to verify that this formula is bound to be true just in case the original truth function $ is true. [If the function under consideration is everywhere false, simply consider the formula (φ&~φ).] In other words, this recipe shows how one can take any truth function whatsoever, and from it construct a formula in SL that will represent that very truth function. SL is thereby shown to be “truth- functionally” or expressively complete.

5. More than that, the formulas produced by this recipe are all in what is called “Disjunctive Normal Form.” A formula in DNF is one which contains no operators other ~,&, and v, and in which no negations operate over anything other than atomic formulas, and no disjunction falls within the scope of any conjunction. Formulas in DNF are basically disjunctions of conjunctions of atomic formulas and their negations. What this shows is that a language employing just (~,&,v) is truth-functionally complete. Moreover, since any formula in SL represents some truth function or other, what this also shows is that every formula in SL has some equivalent in DNF.

6. Now we can do even better than that: Consider an approach that goes “the other way around.” Instead of focusing on those rows of the truth table representation of a function that are true, focus instead upon those rows in which the function is false. In the example above, these are rows 3, 5, and 7. One way of representing the truth function in question is simply to apply the procedure outlined above and then negate the result, saying in effect that the truth-function we want is NOT ANY of the false rows:

~[(Row 3) v (Row 5) v (Row 7)].

But those of you who are familiar with the DeMorgan’s equivalencies1 will see that this formula is equivalent to the following:

[~(Row 3) & ~(Row 5) & ~(Row 7)] or more specifically,

[~(P&~Q&R) & ~(~P&Q&R) & ~(~P&~Q&R)]

In other words, we take the negation of the conjunctive representation of every row on which some target truth function is false, and then conjoin them all together. [If the function is everywhere true, just write down ~(P&~P).] What you get is a formula that looks something like this:

~(row i) & ~(row j) & ~(row k)….for every row where our target truth function is false..

Once again, this is a formula in SL which will be true under just those circumstances in which the original truth function is true. But notice that this formula will only contain the negation and conjunction operators! In other words, (~,&) is expressively complete all by itself.

1 The DeMorgan equivalencies basically state that “Neither A nor B” is equivalent to “Not A and Not B.” and that “Not both A and B” is equivalent to “Either Not A or Not B.” Augustus DeMorgan was a pioneering 19th century mathematical logician who attempted to show how rules of reasoning could be modeled algebraically. In short, DeMorgan’s “rules” direct us how to distribute or to extract negations to and from conjunctions and disjunctions, much as one might distribute or extract factors into or from sums. 7. By the way, we can construct yet another formula equivalent to this one by using the DeMorgan’s rule once again to “drive” the various negations into the conjunctive representation of each row. When we do (and we also use the double negation (or DN) equivalencies to rid ourselves of pesky double negations), we’ll arrive at a formula containing just (~,&,v) as functors, where negations operate only on atomic formulas and no disjunction has any conjunction within its scope. Guess what such a “conjunction of disjunctions of atomic formulas and their negations” are called?

8. Now that we have established the truth-functional completeness of certain sets of truth operators, we are in a position to show that other sets are truth-functionally complete as well. This we can do by showing that a certain set of operators has all the expressive resources of a truth functionally complete one, by “defining” all the operators of the complete set in terms of the other set.

First off, it’s pretty easy to see how to represent or to “define” v in terms of (~, →). We do this by showing that the truth table representation of (φ v ψ) can be replicated by some formula that includes only negations and conditionals. As verified by the following joint truth table, the formula (~φ→ψ) fits the bill:

φ ψ (φ v ψ) (~φ→ψ) T T T T T F T T F T T T F F F F

Similarly, we can straightforwardly define & in terms of (~,v) by using the following Demorgan’s equivalency:

(φ & ψ) : : ~(~φ v ~ψ)

Once again, we can verify this equivalency by means of a truth table. What this means is that we could express the truth function represented by any conjunction by using a combination of negations and disjunction instead. In principle we could “trade in” the conjunction operator without any loss of expressive power, as long we still had the negation and disjunction operations available to us.

9. We were able to conclude in section 6. above that the set (~,&) is truth-functionally complete. Now we know that & can be defined in terms of (~ and v); we also know that ~ can be defined by those same operators because trivially, ~ can define itself. Since the set (~,v) can define all the members of a truth- functionally complete set, we may conclude that the set (~,v) is also expressively truth-functionally. And since v can be defined in terms of negation and the conditional, we also happen to be in position to say the same of (~,→).

10. Perhaps a little more intriguing is the fact that ~ can be defined in terms of (→,⊥). How? [Think of ⊥ as the falsum, a “zero-place” operator that always comes out false.] So given what we know about the set (~,→), what may we conclude from this result?

[As an aside, this is the basic set of truth operators Garson (2006) deploys in his otherwise very fine text on Modal Logic.]

11. So what about sets containing just a single truth-functional operator? Can & or v be truth- functionally complete all by themselves (or even in tandem)? In order for that to be the case, one must be able to express the negation of a formula as a function of just conjunction or disjunction. Think about why you might not be able to do that. What does that suggest about the set of operators containing both (&,v).

The thing that blocks conjunction and disjunction from being expressively complete all by themselves is that a truth conjoined with a truth, and a truth disjoined with a truth, will always come up true. Thus one can never construct a falsehood from just true components. Similarly, one can never construct a truth from just false components. The conditional shares the first problem though not the second. [How can one construct a true conditional out of solely false components?]

12. One might think at this point that to be truth-functionally complete, a set must contain at least two truth functors. But that would be a mistake. Think of the n/and and n/or functions, symbolized by the up arrow and down arrow respectively (↑,↓), aka the Scheffer and Peirce strokes. Notice how, unlike the conjunction and disjunction, a formula that “strokes itself” (my apologies for that!) will switch truth value. Show how each of these individually can be shown to be truth functionally complete by showing that they can individually define negation and either conjunction or disjunction.

13. So far, we have shown how one can go about showing that a given set of truth functors is truth- functionally complete. We’ve also introduced some informal ways of arguing that a set of truth functors will not be truth functionally complete. However, in order to make that informal demonstration more satisfyingly rigorous, we will need to discuss a completely different sort of argumentation…

14. Exercises

Suppose that the operator <++ (backwards arrow slash) is represented by the following truth table:

Φ ψ φ <++ ψ T T F T F F F T T F F F

(1) Define & in terms of <++ and ~. What, then, may we conclude about the expressive adequacy of the set {~,<++}?

(2) Define F (or the falsum) in terms of <++ alone. What then may we conclude about the expressive completeness of the set {→, <++}?

(3) Finally, define & in terms of <++ alone.

(4) How might one go about showing that the truth-functor <++ is not truth functionally complete on its own? Hint: think of some sort of truth function that cannot be expressed by a formula that includes only <++.

(5) Similarly, come up with some reason or hypothesis for thinking that the set (&, V, →) is not truth functionally complete.