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Proof step 5 from proof gcd-mod (previous step)

n0 < n ⊦ gcd(0 * n + n0, n) = gcd(n, n0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n0 < n ⊦ gcd(n0, n) = gcd(n, n0)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%z: ⊦ 0 + m = m; from specification nat-basic2%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-base: ⊦ m * 0 = 0; from specification nat-mult