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Induction Proofs in Ocaml: USING Helper: Lemma plus (S a) b = S (plus a b) b = Zero LHS: plus (S a) b = { case } plus (S a)Zero = { def of plus } match Zero with | Zero -> a | S c -> plus (S a) c = { apply match } (S a) RHS: S (plus a b) = { case } S (plus a Zero) = { def of plus } match Zero with | Zero -> a | S c -> plus (S a) c = { apply match } S (a) Since LHS = RHS, Lemma is proven. Use helper to prove: P3) Prove plus a (plus b c) = plus (plus a b) c

Induction Proofs in Ocaml: USING Helper: Lemma plus (S a) b = S (plus a b) b = Zero LHS: plus (S a) b = { case } plus (S a)Zero = { def of plus } match Zero with | Zero -> a | S c -> plus (S a) c = { apply match } (S a) RHS: S (plus a b) = { case } S (plus a Zero) = { def of plus } match Zero with | Zero -> a | S c -> plus (S a) c = { apply match } S (a) Since LHS = RHS, Lemma is proven. Use helper to prove: P3) Prove plus a (plus b c) = plus (plus a b) c

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Induction Proofs in Ocaml:

USING Helper: Lemma

plus (S a) b = S (plus a b)
    b = Zero


LHS:

plus (S a) b
= { case }
    plus (S a)Zero
= { def of plus }
    match Zero with
    | Zero -> a
    | S c -> plus (S a) c
= { apply match }
    (S a)

RHS:


S (plus a b)
= { case }
    S (plus a Zero)
= { def of plus }
    match Zero with
    | Zero -> a
    | S c -> plus (S a) c
= { apply match }
    S (a)

Since LHS = RHS, Lemma is proven.

Use helper to prove:
P3) Prove plus a (plus b c) = plus (plus a b) c

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