Metatheory - University of Cambridge

1. Induction 9
Now, I hope that it was obvious that the Proposition was true, before we set
about proving it. The purpose of this proof was not, then, to convince you
of its truth. Rather, I gave you this proof primarily in order to illustrate the
Principle of Strong Induction on Length. Moreover, this proof exhibits three
features that are worth mentioning explicitly.
First, even though the result was obvious, it took quite a lot of effort to prove
it. This is fairly typical of metalogical results: showing them with sufficient
rigour can be hard!
Second, we teamed an inductive principle with a recursive definition. This
is fairly typical of metalogical results. Indeed, it is hardly surprising that
induction and recursion should be best of friends. Recursive definitions tell you
how to build objects up, stage-by-stage; proofs by (strong) induction involve
showing that something holds at later stages if it has held at every stage so far.
Third, when we broke the proof down into cases, cases 3–6 were a bit repetitious.
Indeed, all four of our two-place connectives have the same syntactic
behaviour, and that is all we cared about during the course of the proof. So
we could simply have rolled cases 3–6 together in something like the following:
Case 3: A is (B ◁ C ), for some two-place connective ◁ and sentences B and
C . Clearly, both B and C are shorter than A. By supposition, then, B
contains exactly as many instances of ‘(’ as of ‘)’, say i; and C contains
exactly as many instances of ‘(’ as of ‘)’, say j. Hence A contains i+j +1
instances of ‘(’, and i + j + 1 instances of ‘)’.
saving ourselves some time and effort. 6 Generally, if we need to do a proof
by cases – and we often shall – we should pause before ploughing in, to think
about exactly what the relevant cases are.
1.5 Unique readability
I shall close this chapter by proving two more results about the syntax of TFL.
When A and B are expressions, say that B is an initial segment of A
iff A can be obtained by concatenating some (possibly no) symbols to the
(right-hand) end of B. For example:
• ‘¬’ is an initial segment of ‘¬¬A’ and an initial segment of ‘¬(P ∧ Q)’,
but not an initial segment of ‘(¬P → Q)’.
• ‘(P ∧ ¬Q)’ is an initial segment of ‘(P ∧ ¬Q)’, but not an initial segment
of ‘((P ∧ ¬Q) ∨ R)’
Here is a cute result about initial segments.
Lemma 1.6. For any sentence A, the only sentence which is an initial segment
of A is A itself.
Proof. Let A be any sentence, and suppose (for induction) that every shorter
sentence, B, has the property that the only sentence which is an initial segment
of B is B itself (this is the induction property). There are three cases:
6 Note that ‘◁’, like ‘A’, is a symbol of the metalanguage, not the object language. For a
reminder of the distinction, fx C§7.