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Proof step 10 from proof max-bigger (previous step)

((m + 2) * m + m) / 2 < n, maxenc(m + 1, n) < m, ∀ n0, m1. n0 - m1 < n - m → maxenc(m1, n0) < m1 → false ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
n = n1 + m ∧ m ≤ n, ((m + 2) * m + m) / 2 < n, maxenc(m + 1, n) < m, ∀ n0, m1. n0 - m1 < n1 → maxenc(m1, n0) < m1 → false ⊦

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n, m, n1] flags:

Additional information

Used simplifier rules: lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%trans: ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward) from specification nat-basic1%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%two-loup-12: ⊦ (m + n) / 2 < n ↔ m < n; from specification nat-div%ls-01: ⊦ m0 * m < m ↔ m0 = 0 ∧ m ≠ 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%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, m, n1