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Proof step 17 from proof gcd-conq-one (previous step)

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

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [m * m0, m2]

Additional information

Used simplifier rules: ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0; (ws)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward) from specification nat% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m * m0, m2