Give a postcondition for the following algorithm that completely describes how t

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Jun 15th, 2015

Hello!

The answer is i=x+1.

Proof: i starts from 0 and grows by 2, so always remains even. At the last step i is still <x and after this step i>=x. i cannot =x (even i and odd x), so i>x and x>i-2. Subtract (i-2) and obtain

2>x-i+2>0, so odd (x-i+2)=1 and i=x+1.

Please ask if anything is unclear.
Jun 15th, 2015

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