Logic as a tool a guide to formal logical reasoning ( PDFDrive ) 67

k I leave it as exercises for the reader to show that each of these inductive definitions defines the set of all natural numbers.. h22n+ 1 =h2n + 1; They look similar, and yet there is s

k

I leave it as exercises for the reader to show that each of these inductive definitions defines the set of all natural numbers

We now consider the recursive definitions:

1 h1(0) = 1;

2 h1(2n+ 1) = 3n+h1(n);

3 h1(2n) =h1(n) + 1, forn >0

and

1 h2(0) = 0;

2 h2(n+ 2) = 2h2(n) + 3;

3 h2(2n+ 1) =h2(n) + 1;

They look similar, and yet there is something wrong with the second definition What?

First, note thath2(1) =h2(2× 0 + 1) = h2(0) + 1 = 1 Now, let us computeh2(3) On the one hand,h2(3) =h2(1 + 2) = 2h2(1) + 3 = 5 On the other hand,h2(3) =h2(2·

1 + 1) =h2(1) + 1 = 2 Thus, we have obtained two different values, which is definitely bad The problem comes from the fact that the second definition allows for essentially

different generations of the same object, leading us to define the notion of unique gener-ation

1.4.5.5 Unique generation

For the correctness of recursive definitions it should be required that every element of

C( B,F) can be constructed uniquely (up to the order of the steps) Otherwise, definitions

can lead to problems as above More formally, the elements ofC( B,F) are represented

by expressions (terms) built from the elements ofB by applying the operations from F.

Unique generation means that every element ofC( B,F) can be represented by a unique

expression

Example 40

1 The standard definition of natural numbers and Definition 1 above have the unique generation property These can be proved by induction on natural numbers.

2 Definition 2 given above does not satisfy the unique generation property.

3 The set FOR of propositional formulae satisfies the unique generation property, also

known as unique readability property.

4 If the subgroup H of a group G is freely generated by a set of generators B , then it satisfies the unique generation property.

5 If the subgroup H of a group G is not freely generated by a set of generators B , then

it does not satisfy the unique generation property.

Theorem 41 If C( B, F) satisfies the unique generation property then for every mapping

h0:B → X and mappings { F f :X 2n → X | f ∈ F} there exists a unique mapping h:

C( B, F) → X defined by the recursive scheme in Section 1.4.5.

I do not give a proof here, but refer the reader to Enderton (2001)