355hw1solf10

355hw1solf10 - CSE 355 Homework#1 Answers 1.42 a E1 = A B A A A B B A B B A A A B B A B B A B A A A B B A B B A A A B B A B B A B A A A A A B A A A

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CSE 355 Homework #1 Answers 1.42) a) 0 E =   , AB 1 E =                   , , , , , , , , , A A A B B A B A B A B A A B B B A B                                                                                             2 , , , , , , , , , , , , ,... , ... , , , ,... , ... , , , ,..., ... , , ,..., A B A B A B A B A B A B A A A B A A A B A B B B B A B B B A B B B B E AA A A B B A A B B A A B A A B A A A A A A B A A A B BBB B B A B B B A B B B     b) Basis: for 0 E =   , , both of which have one proposition letters and zero operators. Inductive Hypothesis: Assume that     1 po n e n e  for any k e E where 0 kn  Inductive Step: For an expression k e E we have   e v u or   e v u for , k uv E where 0 .
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This note was uploaded on 02/27/2011 for the course CSE 355 taught by Professor Lee during the Fall '08 term at ASU.

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355hw1solf10 - CSE 355 Homework#1 Answers 1.42 a E1 = A B A A A B B A B B A A A B B A B B A B A A A B B A B B A A A B B A B B A B A A A A A B A A A

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