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Question 1. The rank of a formula is defined by recursion on formulas: • rk(A) = 0 for every basic formula. • rk(¬A) = rk(A) + 1. rk(A o B) = max(rk(A), rk(B)) +1 for o e {A, V}. • rk(A → B) = max(rk(A) + 1, rk(B))

Question 1. The rank of a formula is defined by recursion on formulas: • rk(A) = 0 for every basic formula. • rk(¬A) = rk(A) + 1. rk(A o B) = max(rk(A), rk(B)) +1 for o e {A, V}. • rk(A → B) = max(rk(A) + 1, rk(B))

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question 1. The rank of a formula is defined by recursion on formulas: • rk(A) = 0 for every basic formula. • rk(¬A) = rk(A) + 1. • rk(A o B) = max(rk(A), rk(B)) + 1 for o € {A, V}. %3D rk(A → B) = max(rk(A) +1, rk(B)) Prove by induction on formulas that rk(A) < depth(A) for all formulas A.
Transcribed Image Text:Question 1. The rank of a formula is defined by recursion on formulas: • rk(A) = 0 for every basic formula. • rk(¬A) = rk(A) + 1. • rk(A o B) = max(rk(A), rk(B)) + 1 for o € {A, V}. %3D rk(A → B) = max(rk(A) +1, rk(B)) Prove by induction on formulas that rk(A) < depth(A) for all formulas A.
Example of a function defined by recursion on formulas We define the depth of a formula as follows • depth(A) = 0 for every basic formula. depth(¬A) = depth(A) +1. depth(A o B) = max(depth(A), depth(B)) + 1 for ● E {A, V, →}. We also let con(A) be the number of occurrences of logical connectives (A, V, →, -) in the formula A. Lemma 2. con(A) < 2depth(4). Proof. By induction on formulas. Induction base: If A is a basic formula, then con(A) = 0 < 1 = 2º = 2depth(A). Induction step: We have to consider composite formulas. ¬A: By induction hypothesis (i.h.) we have con(A) < 2depth(4). Therefore con(¬A) con(A) + 1 i.h. < 2depth(A) + 1 < 2 * 2depth(A) 2depth(4)+1 2depth(¬A) Ao B where o € {A, V,→}: By induction hypothesis we have con(A) < 2depth(A) and con(B) < 2depth(B). con(A o B) con(A) + con(B) + 1 < con(A) +1+ con(B) +1 i.h. < 2depth(A) + 2depth(B) < 2* 2max(depth(A),depth(B)) 2depth(AoB) This completes the proof. ||||
Transcribed Image Text:Example of a function defined by recursion on formulas We define the depth of a formula as follows • depth(A) = 0 for every basic formula. depth(¬A) = depth(A) +1. depth(A o B) = max(depth(A), depth(B)) + 1 for ● E {A, V, →}. We also let con(A) be the number of occurrences of logical connectives (A, V, →, -) in the formula A. Lemma 2. con(A) < 2depth(4). Proof. By induction on formulas. Induction base: If A is a basic formula, then con(A) = 0 < 1 = 2º = 2depth(A). Induction step: We have to consider composite formulas. ¬A: By induction hypothesis (i.h.) we have con(A) < 2depth(4). Therefore con(¬A) con(A) + 1 i.h. < 2depth(A) + 1 < 2 * 2depth(A) 2depth(4)+1 2depth(¬A) Ao B where o € {A, V,→}: By induction hypothesis we have con(A) < 2depth(A) and con(B) < 2depth(B). con(A o B) con(A) + con(B) + 1 < con(A) +1+ con(B) +1 i.h. < 2depth(A) + 2depth(B) < 2* 2max(depth(A),depth(B)) 2depth(AoB) This completes the proof. ||||
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