, Proof" Suppose that 'p' is prime and 'p' divides 'a*b'. If 'not p | a ' then 'gcd(a,p)=1' ; and therefore, by theorem 24, there are 'x' and 'y' such that 'x*a + y*p = 1' or 'x*a*b + y*p*b = b'. We conclude that 'p|b' because 'p|a*b' and 'p|p*b
, { assume n = 0 deduce gcd(n, m) = m 5
, Assume that m and n are relatively prime integers
, If n = 0 then m = 1 because m and n are coprime
, Prove that '-5 <= |x+2| -|x-3| <=5'Proof" We proceed by the case analysis. Case: 'x <= -2'. In this case, '|x+2| -|x-3| = -(x+2) -(-(x-3)) = -5'. Therefore
, We give MathAbs in verbatim to show exactly how the MathAbs' output look like: 1. "Theorem" let x:Integer let y:Integer assume positive(x) land positive(y) land (odd(x) land odd(y)) show even
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