PHIL 422 Advanced Logic Inductive Proof 1. Preamble: One of the most powerful tools in your meta-logical toolkit will be proof by induction. Just about every significant meta-logical result relies upon inductive proof in some way. So you will need to become extremely familiar with how inductive proofs are constructed 2. Let s start with an example from mathematics. Note the following series of sums: 1=1 1+3=4 1+3+5=9 1+3+5+7=16 1+3+5+7+9=25 Notice a pattern? In the first 5 cases at least, the sum of the first n odd numbers seems to be n 2. Does this pattern continue? It seems so; the sum of the first 6 odd numbers is 25+11=36. And the sum of the first 7 would be 36+13=49. Surely, this can t just be pure coincidence! But how can we assure ourselves that there isn t some weird number way out there for which this result doesn t hold, or some place out on the number line where this pattern stops. 3. Inductive proof provides us with this assurance. In inductive proofs, we show that some hypothesis holds for the earliest members of a series, and then we use that result to show that it must therefore hold for all later ones as well. The procedure: Step 1. We start by framing a hypothesis about all of the members of a class or sort. ( Every S is a Φ. ) Step 2. We then specify a way in which all the members of that class may be ordered. ( All of the S s may be ordered in terms of.. ) Here are two examples: (i) The set of all natural numbers (0, 1, 2, 3, etc.) can be ordered according to their magnitude. In this case, we can say that the ordering is complete or total, because for every pairs of numbers, either one is greater than, or less than, the other (there are no ties ). (ii) The set of all sentences in our logical language can be ordered according to
the number of truth-functors that they contain (you may think of this as their length ). Unlike the case above, this ordering is only partial, since distinct sentences might have an equal number of truth factors. Step 3. We then show that this hypothesis holds for the first elements of that ordering (the base cases). ( The first or earliest S s are Φ. ) For example, for an arithmetical generalization, we show that the hypothesis holds for 0 or for 1 (depending upon whether our hypothesis concerns the natural numbers or the counting numbers). For a generalization about logical formulas, we show that the hypothesis holds of atomic sentences. In many cases, this will be trivial. Step 4. We then consider an arbitrary element of a series, and show that, on the assumption that the hypothesis applies to all of the elements earlier in the series (called the inductive hypothesis), the hypothesis must apply to that arbitrary element as well. This is sometimes called the inductive step of the argument. ( If all the earliers S s are Φ, then the subsequent S s must also be Φ ) 4. Let s return to our mathematical example above, and apply the steps of inductive proof. Step 1. Our hypothesis was as follows: The sum of the first n odd numbers is equal to n 2. Step 2. We shall order all of the numbers in the obvious way by their magnitude. Step 3. Proof of the base case: We simply observe or verify that the hypothesis holds for the first case. Namely, the sum of the first 1 odd numbers (1) is equal to 1 2. Step 4. Proof of the inductive step: Take an arbitrary number x, and show that the sum of the first x odd numbers is equal to x 2, on the assumption that the hypothesis holds for any number less than x. This will takes some ordinary algebra (sorry!). But first observe that the sum of the first x odd numbers will be equal to the sum of the first x-1 odd numbers plus the x th odd number. Now that x th odd number will be equal to 2x-1 (you can easily verify this). So the sum of the first x odd numbers will be equal to the sum of the first x-1 odd numbers plus 2x-1. Now observe that x-1 is obviously less than x, and so our inductive assumption
must hold of it. That means that the sum of the first x-1 odd numbers is assumed to be (x-1) 2. We now plug in those two values and do the algebra. The sum of the first x odd numbers is now understood to be: The sum of the first x-1 odd numbers + the x th odd number = (x-1) 2 + (2x-1) = (x 2-2x +1) +(2x-1) = x 2 Which is exactly what we wanted to show! Our inductive proof of the hypothesis is now complete! 5. An exercise: try it for yourself! Construct a similar inductive proof establishing that 1+2+3+ +(n)= n(n+1)/2. 6. So why do we care? We know how to establish the expressive or truth-functional completeness of a set of truth functors, but until now we had no formal way of establishing the expressive incompleteness of a set of functors. Now we do, once we begin to apply inductive proof to sentences of our logical language. A Second Example: The expressive limitation of {&, v} Step 1: Our hypothesis is that if we restrict our logical language to atomic sentences and compounds formed by conjunction and disjunction (&, v), then we could never construct a formula that takes on the value true when all of its atomic components are false. Step 2: This time, we ll order all of the relevant formulas by their size (as measured by the number of functors they contain). Step 3: Atomic formulas. Trivially, atomic formulas are false when they are false. The hypothesis clearly holds. Step 4: Given our restriction, we only have two sorts of compounds to consider, conjunctions and disjunctions: First, suppose a compound formula is a conjunction, and that the inductive hypothesis applies to any formula shorter than it. Now, clearly, each of its conjuncts is shorter than it, so they must take on the value false whenever all of their atomic components are false. But that means that when all of the atomic components of our conjunction are false, both of its conjuncts must be false. And that means that our original conjunction must be false as well.
An entirely parallel line of reasoning applies in the case that a compound formula is a disjunction (check it for yourself!). Now there is a somewhat interesting upshot of this inductive proof. It establishes that conjunctions and disjunctions alone will not be able to express any truth function that takes on the value true when all of its atomic components are false. In other words, it establishes that the set of truth-functors {&,v} cannot be truth-functionally complete, no matter how hard you try or creative you get. 7. A Third Example: The expressive limitation of the conditional Step 1: Our hypothesis is that no formulas containing only {P, Q, } has a truth table with more than two F s (considered as a 2-place truth function) Step 2: Once again, we ll order all of the relevant formulas by their size (as measured by the number of functors they contain). Step 3: Atomic Formulas: The only atomic formulas to consider here are P and Q. Considered as a 2-place truth function (whose truth table has four rows), its truth table will have exactly two F s. Step 4: Compound formulas: Since we re restricting our language, we only have to consider compound cases which are conditionals. And since the consequent of any conditional is clearly shorter than the conditional itself, the inductive hypothesis holds for it. So we can safely conclude that our conditional s consequent has at most two rows which are false. But since a conditional can be false only when its consequent is false, we may conclude that the conditional itself can be false on at most two rows. The upshot: note that conjunction is a truth function that is false on three rows of its truth table. That means that our inductive proof shows that we cannot express or define the conjunction function by means of the conditional alone. 8. Some exercises: (1) Define conjunction in terms of the conditional and the biconditional (2) Now think of a way to use that result, in combination with the inductive proof above, to establish that the conditional alone cannot define the biconditional. (3) Show that the set {, ~) is also expressively incomplete. To do this, you will need to identify some property that all negations and biconditionals have, which would prevent them from being able to express certain kinds of truth functions. Then use an inductive proof to show that every compound formula constructed from biconditionals and
negations must have this property. (4) Observe the following striking pattern: 1 3 = 1 1 3 +2 3 = 9 1 3 +2 3 +3 3 =36 1 3 +2 3 +3 3 +4 3 =100 1 3 +2 3 +3 3 +4 3 +5 3 =225 In other words, the sum of the cubes of the first n numbers is equal to the sum of the first n numbers squared. [1 3 +2 3 +3 3 + n 3 = (1+2+3+ n) 2 ] Now, using the result of section 5. above (a result that itself required an inductive proof), provide an inductive proof establishing that this general pattern holds.