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Proof step 6 from proof modp (previous step)

m0 * m1 + n0 < m0 * (n - 1), m0 * n ≤ (m0 * m1 + m0) + n0, m1 + 1 < n, n0 < m0, ∀ n1, m, n2. ¬ m1 < n1 ∨ ¬ m * n1 ≤ m * m1 + n2 ∨ ¬ n2 < m ⊦ n - 1 = 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] → [n, m2] flags:

Additional information

Used simplifier rules: mult-base: ⊦ m * 0 = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%dist: ⊦ m0 * (m + n) = m0 * m + m0 * n;%one: ⊦ m0 * 1 = m0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat%lez: ⊦ 0 ≤ m; from specification nat%l-num-06: n0 < n ⊦ ¬ m + (* n) < (* n0); from specification nat%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m2