Metatheory - University of Cambridge

1. Induction 8
sentences. But things will go easier for us if we introduce yet another inductive
principle that relates more directly to TFL:
Principle of Strong Induction on Length. Let ϕ be any property. Suppose
that, for any sentence, A, if every shorter sentence has ϕ, then A has ϕ. Then
every sentence has ϕ.
This principle follows directly from (ordinary) strong induction:
Proof. Suppose that, for any sentence, A, if every shorter sentence has ϕ, then
A has ϕ. We aim to show that this entails that every sentence has ϕ (thereby
vindicating the Principle of Strong Induction on Length).
To do this, we shall define a new property of the numbers, ψ, thus:
every sentence of length x has ϕ
Pick any arbitrary number, n, and suppose for (strong) induction that every
number less than n has ψ. That is, every sentence of length < n has ϕ. So,
by supposition, every sentence of length n has ϕ. So n has ψ. It follows, by
strong induction, that every number has ψ.
Since every sentence has some number as its length, it now follows that
every sentence has ϕ, as required.
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So, armed with the notion of the length of a sentence, and the Principle of
Strong Induction on Length, we can now prove Proposition 1.5:
Proof of Proposition 1.5. Let A be any sentence, and suppose for (strong) induction
(on length) that every shorter sentence has exactly as many instances
of ‘(’ as of ‘)’. We shall show that A has exactly as many instances of ‘(’ as of
‘)’, by considering all the different forms that A might have. (These six cases
are given to us by the recursive definition of the notion of a sentence; for a
reminder of the definition, see fx C§6.)
Case 1: A is atomic. Then A contains no brackets, so it has exactly as
many instances of ‘(’ as of ‘)’.
Case 2: A is ¬B, for some sentence B. Clearly, B is shorter than A. By
supposition, then, B contains exactly as many instances of ‘(’ as of ‘)’.
Hence A does too.
Case 3: A is (B ∧ C ), for some 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 ‘)’.
Case 4: A is (B ∨ C ), for some sentences B and C . Exactly similar to
case 3.
Case 5: A is (B → C ), for some sentences B and C . Exactly similar to
case 3.
Case 6: A is (B ↔ C ), for some sentences B and C . Exactly similar to
case 3.
There are no other forms that A could take. Hence, for any sentence, A, if
every shorter sentence has exactly as many instances of ‘(’ as of ‘)’, then so
does A. The result now follows by strong induction on length.
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