536_solution3 - Homework #3 Solution Answer We get Answer...

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Unformatted text preview: Homework #3 Solution Answer We get Answer The predicate is False for z=0. Hence, does not satisfy the given predicate. Answer Answer Answer Let P(x) = Let Then Now consider To disprove Choosing then: , we only need one value c where Hence, does not imply that holds. Answer Answer Answer If n>0 then s:=0; k:=0; b[0]:=0; while k<n do k:=k+1; b[k]:=k; s:=s+k od k:=k+1; else skip fi Answer Answer Note that, is the key. Answer a) b) ...
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This note was uploaded on 04/17/2011 for the course CS 536 taught by Professor Cs536 during the Fall '08 term at Illinois Tech.

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