PHIL 2610 Exercise Week 3

Consider the language of propositional logic, with atomic propositional letters P,Q,R, . . .,and compound w!s built up from them in the usual way using the connectives ¬,! and".

Let a Kleene strong model consist of an assignment to each atomic proposition of a truthvalue (1, # or 0), with the truth values of compound formulas on a model being given bythe clauses:

• [! ! "] = [!] ! ["]

• [! " "] = [!] " ["]

• [¬!] = [!]!

where !," and ! on the right hand side are the operations in the Kleene strong algebraof truth values.

Let I1 and I2 be Kleene strong models. Write I1 $ I2 to mean that for every atomicpropositional letter X:

1. If X has value 1 on I1, then X has value 1 on I2

2. If X has value 0 on I1, then X has value 0 on I2

Show by induction on complexity of formulas (see attached explanation) that if I1 $I2, then these two conditions hold for every formula whatsoever (i.e. not just atomicformulas), i.e. where ! is any formula:

1. If ! has value 1 on I1, then ! has value 1 on I2

2. If ! has value 0 on I1, then ! has value 0 on I2

Proof by Induction

If we want to show that a certain property P holds of every one of a certain bunch ofthings, it is often useful to prove this by induction (aka mathematical induction). Suchproofs proceed as follows:

(A) Enumeration:We assign to each object a number 1, 2, 3, . . .

(B) Base case:We prove that property P holds of the thing(s) numbered 1.

(C) Induction step:We prove that the property holds of the thing(s) numbered n+1, on the assumption thatthe property holds of all the thing(s) numbered 1 through n. This assumption is calledthe inductive hypothesis.

Let’s think through how these steps do indeed establish that every object in questionhas the property P . By step (C), on the assumption that all objects numbered 1 havethe property P , it follows that all objects numbered 2 have the property P . But bystep (B), all objects numbered 1 do have the property P . So it follows that all objectsnumbered 2 have the property P . Now by step (C) again, on the assumption that allobjects numbered 1 and 2 have the property P , it follows that all objects numbered 3have the property P . But we have already established that all objects numbered 1 and allobjects numbered 2 have the property P : so it follows that all objects numbered 3 havethe property P . Now by step (C) again, on the assumption that all objects numbered 1and 2 and 3 have the property P , it follows that all objects numbered 4 have the propertyP . But we have already established that all objects numbered 1 and all objects numbered2 and all objects numbered 3 have the property P : so it follows that all objects numbered4 have the property P . In general, whatever number n we pick, we can establish by thisstyle of reasoning that all objects numbered n have P . But every object has been givensome number n—that is step (A)—so it follows that every object has P .

There is a variant form of proof by induction where, in the induction step, we prove thatthe property holds of the thing(s) numbered n+ 1, on the assumption that the propertyholds of the thing(s) numbered n (rather than 1 through n). Often we can get by withthis weaker inductive hypothesis (‘weaker’ in that it is implied by, but does not imply,the original inductive hypothesis). Sometimes the terms ‘strong induction’ or ‘course-of-values induction’ are used for induction with the stronger inductive hypothesis; sometimesthe term ‘weak induction’ is used for induction with the weaker inductive hypothesis.

Another common variation involves numbering the objects from 0 rather than from 1 instep (A)—i.e. 0, 1, 2, 3, . . . rather than 1, 2, 3, . . .. In this case, in step (B) we prove thatproperty P holds of the thing(s) numbered 0 (rather than 1), and in step (C) the (strong)inductive hypothesis assumes that P holds of all the thing(s) numbered 0 through n (not1 through n).

The key to a successful inductive proof often lies in step (A)—in finding a way of assigningnumbers to the objects under consideration which facilitates establishing steps (B) and

(C). When the objects under consideration are w!s—when we want to show that all ofsome bunch of w!s have a certain property—one way of assigning numbers to them whichis often very useful is this: the number that we assign to a w! is the number of logicaloperators (i.e. connectives and/or quantifiers) that it contains. So atomic w!s (e.g. Pa,b = c) get number 0; w!s with one connective or quantifier (e.g. (Pa ! Rb), ¬b = c,(%x)Rx) get number 1; and so on. The number of logical operators contained in a w!! is often called the complexity of !; a proof by induction where the objects in questionare w!s, and these w!s are assigned numbers in step (A) according to their complexityin the way just described, is often called a proof by induction on complexity of formulas.Such proofs are common in logic.