1.11 Inductive definitions and structural induction
- 1. Introduction to set theory and to methodology and philosophy of mathematics and computer programming Inductive definitions and structural induction An overview by Jan Plaza c 2017 Jan Plaza Use under the Creative Commons Attribution 4.0 International License Version of February 14, 2017
- 2. successor(n) = n + 1 Natural numbers can be constructed from 0 by repeated applications of successor. To prove that every natural number has a property P, one can use mathematical induction. Sets of other mathematical objects can be defined by repeated applications of some constructs to certain base objects – these are inductive definitions . To prove that every object in such a set has a property P, one can use structural induction .
- 3. Definition Let Prop be a set of propositional symbols. The set of propositional formulas over Prop , denoted Form(Prop) , is defined as follows in an inductive definition. 1. , ⊥ ∈ Form(Prop). 2. If p ∈ Prop, then p ∈ Form(Prop). 3. If A ∈ Form(Prop), then ¬A ∈ Form(Prop). 4. If A, B ∈ Form(Prop), then (A ∧ B), (A ∨ B), (A → B) ∈ Form(Prop). 5. Nothing belongs to Form(Prop) unless it can be constructed using the preceding rules.
- 4. Definition Let Prop be a set of propositional symbols. The set of propositional formulas over Prop , denoted Form(Prop) , is defined as follows in an inductive definition. 1. , ⊥ ∈ Form(Prop). Base clause 2. If p ∈ Prop, then p ∈ Form(Prop). Base clause 3. If A ∈ Form(Prop), then ¬A ∈ Form(Prop). Inductive clause 4. If A, B ∈ Form(Prop), then (A ∧ B), (A ∨ B), (A → B) ∈ Form(Prop). Inductive clause 5. Nothing belongs to Form(Prop) unless it can be constructed using the preceding rules. Restriction clause
- 5. Definition Let Prop be a set of propositional symbols. The set of propositional formulas over Prop , denoted Form(Prop) , is defined as follows 1. , ⊥ ∈ Form(Prop). 2. If p ∈ Prop, then p ∈ Form(Prop). 3. If A ∈ Form(Prop), then ¬A ∈ Form(Prop). 4. If A, B ∈ Form(Prop), then (A ∧ B), (A ∨ B), (A → B) ∈ Form(Prop). 5. Nothing belongs to Form(Prop) unless it can be constructed using the preceding rules. Notice the parentheses in (A ∧ B), (A ∨ B), (A → B).
- 6. Proposition Let Prop be a set of propositional symbols. Every formula X from Form(Prop) contains equal number of left and right parentheses. Proof by structural induction Base case: X is . has equal number of left and right parentheses, namely 0. Base case: X is ⊥. ⊥ has equal number of left and right parentheses, namely 0. Base case: X is a propositional symbol p from Prop. p has equal number of left and right parentheses, namely 0. Inductive case: X is ¬A, for some A ∈ Form(Prop). By inductive hypothesis, A has equal number of left and right parentheses. X, which is ¬A, has the same numbers of left and right parentheses as A. So X has equal number of left and right parentheses.
- 7. Inductive case: X is (A ∧ B), for some A, B ∈ Form(Prop). By inductive hypothesis, A has equal number of left and right parentheses, say m, and B has equal number of left and right parentheses, say n. Then, X, which is (A ∧ B), has the m + n + 1 left parentheses, and the same number of right parentheses. Inductive case: X is (A ∨ B), for some A, B ∈ Form(Prop). By inductive hypothesis, A has equal number of left and right parentheses, say m, and B has equal number of left and right parentheses, say n. Then, X, which is (A ∨ B), has the m + n + 1 left parentheses, and the same number of right parentheses. Inductive case: X is (A → B), for some A, B ∈ Form(Prop). By inductive hypothesis, A has equal number of left and right parentheses, say m, and B has equal number of left and right parentheses, say n. Then, X, which is (A → B), has the m + n + 1 left parentheses, and the same number of right parentheses.
- 8. Exercise Prove by structural induction that for every A ∈ Form(Prop), if A does not involve , ⊥ then the number of left parentheses in A is one less than the number of occurrences of propositional symbols in A.
- 9. Definition Let A be a propositional formula. The set of subformulas of A , denoted subformulas(A) , is defined as follows in an inductive definition. 1. A ∈ subformulas(A). 2. If (B ∧ C) ∈ subformulas(A), then B, C ∈ subformulas(A). 3. If (B ∨ C) ∈ subformulas(A), then B, C ∈ subformulas(A). 4. If (B → C) ∈ subformulas(A), then B, C ∈ subformulas(A). 5. If ¬B ∈ subformulas(A), then B ∈ subformulas(A). 6. Nothing belongs to subformulas(A) unless it can be constructed using the preceding rules.
- 10. Definition Let A be a propositional formula. The set of subformulas of A , denoted subformulas(A) , is defined as follows in an inductive definition. 1. A ∈ subformulas(A). Base clause 2. If (B ∧ C) ∈ subformulas(A), then B, C ∈ subformulas(A). Inductive clause 3. If (B ∨ C) ∈ subformulas(A), then B, C ∈ subformulas(A). Inductive clause 4. If (B → C) ∈ subformulas(A), then B, C ∈ subformulas(A). Inductive clause 5. If ¬B ∈ subformulas(A), then B ∈ subformulas(A). Inductive clause 6. Nothing belongs to subformulas(A) unless it can be constructed using the preceding rules. Restriction clause
- 11. Definition Let A be a propositional formula. The set of subformulas of A , denoted subformulas(A) , is defined as follows in an inductive definition. 1. A ∈ subformulas(A). 2. If (B ∧ C) ∈ subformulas(A), then B, C ∈ subformulas(A). 3. If (B ∨ C) ∈ subformulas(A), then B, C ∈ subformulas(A). 4. If (B → C) ∈ subformulas(A), then B, C ∈ subformulas(A). 5. If ¬B ∈ subformulas(A), then B ∈ subformulas(A). 6. Nothing belongs to subformulas(A) unless it can be constructed using the preceding rules. Exercise: Let A be the formula (p → (q ∨ ¬p)). Find the set subformulas(A).
- 12. Definition Let A be a propositional formula. The set of subformulas of A , denoted subformulas(A) , is defined as follows in an inductive definition. 1. A ∈ subformulas(A). 2. If (B ∧ C) ∈ subformulas(A), then B, C ∈ subformulas(A). 3. If (B ∨ C) ∈ subformulas(A), then B, C ∈ subformulas(A). 4. If (B → C) ∈ subformulas(A), then B, C ∈ subformulas(A). 5. If ¬B ∈ subformulas(A), then B ∈ subformulas(A). 6. Nothing belongs to subformulas(A) unless it can be constructed using the preceding rules. Exercise: Let A be the formula (p → (q ∨ ¬p)). Find the set subformulas(A). Answer: {(p → (q ∨ ¬p)), p, (q ∨ ¬p), q, ¬p}