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Proof step 3 from proof gcd-mod (previous step)

m = m0 * n + n0 ∧ n0 < n ⊦ gcd(m, n) = gcd(n, n0), n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n0 < n ⊦ gcd(m0 * n + n0, n) = gcd(n, n0)

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; from specification nat-basic1