lecture4_W10

# lecture4_W10 - CS245 Winter 2010 Lecture 4 Shai Ben-David...

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CS245 - Winter 2010, Lecture 4 Shai Ben-David Proofs by structural Induction In the previous lecture we proved that one can apply proofs by structural induc- tion to show that a property holds for all elements of a set deﬁned as I ( A,P ). Namely, Theorem [Proof by structural induction] : Let T denote any property of elements of a domain set X . For any sets A and P , if 1. T holds for all members of A . 2. T is retained by the operations of P . Namely, for every operation, p P and for every x 1 ,...,x m X , if T holds for each of x 1 ,...,x m then T holds for p ( x 1 ,...,x m ) (note that the ’ m ’ here denotes the number of of inputs the operation p requires. So if p is operates on single elements, like the ¬ operation, m = 1 , if p takes pairs of elements as input, like the connective, then m = 2 , etc.). Then T holds for all members of I ( A,P ) . A proof by structural induction is quite similar to the familiar ”proofs by induction” for the natural numbers. That is, we have a base step , in which we show that the property in question applies to all members of the core set, A (this step is analogous to the base step on proofs by induction for natural numbers, in which one shows that the property holds for the number 0). The second step in a proof of this type is the inductive step . In the inductive step one has to show that the property is preserved under the application of the operators in P . In other words, one need to show that if any elements b 1 ,...b

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lecture4_W10 - CS245 Winter 2010 Lecture 4 Shai Ben-David...

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