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Proof step 15 from proof gcd-conq-one (previous step)

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

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: zero-06: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0); from specification nat-mult%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%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m * m0, m2