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

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

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Understanding Propositional Logic 43 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: h1 (0) = 1; h1 (2n + 1) = 3n + h1 (n); h1 (2n) = h1 (n) + 1, for n > and h2 (0) = 0; h2 (n + 2) = 2h2 (n) + 3; h2 (2n + 1) = h2 (n) + 1; They look similar, and yet there is something wrong with the second definition What? First, note that h2 (1) = h2 (2 × + 1) = h2 (0) + = Now, let us compute h2 (3) On the one hand, h2 (3) = h2 (1 + 2) = 2h2 (1) + = On the other hand, h2 (3) = h2 (2 · + 1) = h2 (1) + = 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 generation 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 of C(B, F) are represented by expressions (terms) built from the elements of B by applying the operations from F Unique generation means that every element of C(B, F) can be represented by a unique expression Example 40 The standard definition of natural numbers and Definition above have the unique generation property These can be proved by induction on natural numbers Definition given above does not satisfy the unique generation property The set FOR of propositional formulae satisfies the unique generation property, also known as unique readability property If the subgroup H of a group G is freely generated by a set of generators B , then it satisfies the unique generation property 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 {Ff : 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 not give a proof here, but refer the reader to Enderton (2001)

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