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
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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.
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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