Quiz3_Review_Solution

Quiz3_Review_Solution - Solution to Questions from Quiz 3...

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Unformatted text preview: Solution to Questions from Quiz 3 Review (Fix bugs in 1e and 1f) Sample Problems 1. Try expanding the following proof outlines. If you nd any predicate logic obligations, list them. a. {} y := 0; x := f ( y ) { x ! f ( y )} { f (0) ! f (0)} y := 0; { f ( y ) ! f ( y )} x := f ( y ) { x ! f ( y )} b. {} x := 0; y := f (0) { x ! f ( y )} {0 ! f ( f (0))} {0 ! f ( f (0))} x := 0; { x ! f ( f (0))} y := f (0) { x ! f ( y )} I think in practice wed leave out the 0 ! f ( f (0)) condition and 0 ! f ( f (0)) ! f ( f (0)) obligation. c. {} z := z + e ; y := y +1 { y = z } {( y +1) = z + e } z := z + e ; {( y +1) = z } y := y +1 { y = z } d. {} z := z + e ; y := y +1 { y > z } {( y +1) > z + e } z := z + e ; {( y +1) > z } y := y +1 { y > z } [Sorry, I forgot to make (c) and (d) more di erent as problems!] e. {} if x > 0 then x := y { y > 0} [Dont forget to add the else skip clause] { x ! 0 y > 0} [Simpli es the condition that follows] {( x > 0 " x > 0) # ( x ! 0 ! y...
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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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Quiz3_Review_Solution - Solution to Questions from Quiz 3...

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