Lecture3 - Lecture 3 Inductively Defined Sets and...

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Lecture 3 Inductively Defined Sets and Definition by Recursion Review Inductively defined sets are generated from operations on base elements. Examples: n : Natural numbers-generated : from 0 by operation of ‘_ + 1’ 1 0 PROP - generated from propositional ( (p 1 ˮ p 3 ) p 8 ) symbols and by operations of ˭ , ˮ , ± ¬ ±  ² p 1 ˮ p 3 ) p 8 p 1 p 3
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Lecture 3 Inductively Defined Sets and Definition by Recursion Review Inductively defined sets are generated from operations on base elements. We prove that a property holds of an inductive set by showing that 1) it holds of the base elements (base case). 2) if holds of given elements, then it holds of all elements that can be generated from them (inductive step). This is mathematical induction. The topic for today is definition by recursion . It is a way to define a function on an inductively defined set. But before getting to it, we’ll talk a little more about inductive proofs about PROP .
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Mathematical induction with PROP Claim : all members of PROP have an even number of brackets. Proof : Base case Both p i and have 0 brackets. 0 is an even number. So p i and have an even number of brackets. Inductive step Case ( φ ˭ ψ ) . Assume for the inductive hypothesis that arbitrary propositions φ and ψ have an even number of brackets. Suppose in particular that φ has 2n brackets, and ψ has 2m brackets. Then ( φ ˭ ψ ) has 2 + 2n + 2m = 2(1 + n + m) brackets. Thus ( φ ˭ ψ ) has an even number of brackets. A similar argument works for the cases ( φ ˭ ψ ), ( φ ˮ ψ ), ( φ ψ ) and ( φ  ψ ). Case ( ¬ φ ). Assume for the inductive hypothesis that φ has an even number of brackets—i.e. assume it has 2n brackets. Then ( φ ) has 2 + 2n = 2(1 +n) number of brackets—i.e. an even number of brackets.
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Mathematical induction with PROP Claim : all members of PROP . Proof : Base case Both p i and have 0 brackets. 0 is an even number. So p i and . Inductive step Case ( φ ˭ ψ ) . Assume for the inductive hypothesis that arbitrary propositions φ and ψ have an even number of brackets . Suppose in particular that φ has 2n brackets, and ψ has 2m brackets. Then ( φ ˭ ψ ) has 2 + 2n + 2m = 2(1 + n + m) brackets. Thus ( φ ˭ ψ ) . A similar argument works for the cases ( φ ˭ ψ ), ( φ ˮ ψ ), ( φ ψ ) and ( φ  ψ ). Case ( ¬ φ ). Assume for the inductive hypothesis that φ has an even number of brackets —i.e. assume it has 2n brackets. Then ( φ ) has 2 + 2n = 2(1 +n) number of brackets—i.e. .
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PROP Claim : . Proof : Base case . Inductive step Case ( φ ˭ ψ ) . Assume for the inductive hypothesis that arbitrary propositions φ and ψ have an even number of brackets . Case (
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Lecture3 - Lecture 3 Inductively Defined Sets and...

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