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Proof step 6 from proof modls-03 (previous step)

m0 * n + n1 < (m - 1) * n + n2, m0 + 1 < m, n1 < n, m * n + n2 ≤ (n * m0 + n) + n1, n2 < n ⊦

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, m1] flags:

Additional information

Used simplifier rules: 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%l: ⊦ m + n < m + n0 ↔ n < n0; 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%l-num-06: n0 < n ⊦ ¬ m + (* n) < (* n0); from specification nat%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m1