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Proof step 14 from proof gcd-conq-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
gcd(m, n) = 1, m * m0 = m1 * n + 1, 1 ≤ m * m0 ⊦ n * m1 = m * m0 - 1, n = 0, n = 1, m = 0, m0 = 0

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%zero-01: ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0; from specification nat-mult%not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward) from specification nat