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Structural Induction CS 611 Structural Induction • Useful for proving properties about languages. • Useful for proving properties about algorithms defined on languages. • A generalization of the induction you know and love. Induction on the Naturals 0 To prove : n N .P(n) prove 1. P(0) 2.n N .P(n) P(n 1) P is true of the basis case and if P is true of a number, then n n+1 P is true of that number 1. And "1" is how you get new numbers. Example 0 n n+1 Prove: every natural number is odd or even. Base: 0 is even. Induction: if n is odd or even then n+1 is odd or even. Suppose n is odd, then n+1 is even Suppose n is even then n+1 is odd. Induction on Aexp var const + Suppose I want to prove a property of all arithmetic expressions. Like “eval (aexp) is finite” for some evaluation function eval. Aexp has 2 atoms and 2 constructors. Aexp Aexp - Aexp Aexp So induction on Aexp has ___ base cases and ___ inductive steps? What is the induction theorem for arithmetic expressions? Induction on Aexp var const + Aexp Aexp - Aexp Aexp To prove b Aexp .P (b), prove : 1.P(const) 2.P ( var) 3.a, b Aexp .P(a ) P(b) P(a b) 4.a, b Aexp .P(a ) P (b) P(a - b) Example Eval (aexp a) if a = var then val(a) if a = const then a if a = (b+c) then eval(b)*eval(c) if a = (b-c) then eval(b)/eval(c) Prove: forall a:aexp. eval(a) terminates Base 1: eval(var) terminates Base 2: eval(const) terminates Ind1: suppose eval(b) and eval(c) terminate prove eval (b+c) termiantes Ind2: suppose eval(b) and eval(c) terminate prove eval(b-c) terminates Example Eval (aexp a) if a = var then val(a) if a = const then a if a = (b+c) then eval(b)*eval(c) if a = (b-c) then eval(b)/eval(c) Prove: forall a:aexp. eval(a) = correct_val(a) Base 1: eval(var) = correct_val(var) Base 2: eval(const) = correct_val(const) Ind1: suppose eval(b) = correct_eval(b) and eval(c) = correct_eval(c) prove eval(b+c) = correct_eval(b+c) Ind2: suppose eval(b) = correct_eval(b) and eval(c) = correct_eval(c) prove eval(b-c) = correct_eval(b-c) Example Eval (aexp a) if a = var then val(a) if a = const then a if a = (b+c) then eval(b)*eval(c) if a = (b-c) then eval(b)/eval(c) Prove: forall a:aexp. eval(a) is finite Base 1: eval(var) is finite Base 2: eval(const) is finite Ind1: suppose eval(b) is finite and eval(c) is finite prove eval(b+c) is finite Ind2: suppose eval(b) is finite and eval(c) is finite prove eval(b-c) is finite Reminders • Reverse engineer proof of Thm 3.1 to scratch work for Monday. • Start on proofs of Thms 3.2, 3.3, probably for Tuesday. • Thm 3.4 and Cor 3.2 likely for Friday • Paper 1 on Friday • Cancel Godel encoding part of hwk.