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Proof step 16 from proof gcd-conq-one (previous step)

m * m0 = m1 * n + 1, gcd(m, n) = 1, 1 ≤ m * m0 ⊦ n * m1 = m * m0 - 1, n = 0, n = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
m * m0 = m1 * n + 1, gcd(m, n) = 1, 1 ≤ m * m0 ⊦ n * m1 = m * m0 - 1, m = 0, m0 = 0, n = 0, n = 1

Rule argument

Additional information

Used simplifier rules: mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0; (ws)%not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward) from specification nat