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Proof step 3 from proof max-def-01 (previous step)

⊦ (n * (n + 1)) / 2 ≤ m ∧ m < ((n + 1) * (n + 2)) / 2 → maxenc(0, m) = n

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-pair inst: name: max-def seq: ⊦ (n * (n + 1)) / 2 ≤ m ∧ m < ((n + 1) * (n + 2)) / 2 ∧ (m0 * (m0 + 1)) / 2 ≤ m → maxenc(m0, m) = n subst: [n, m, m0] → [n, m, 0] flags: rewrite, current

Additional information

Used simplifier rules: rsucc-02: ⊦ m * (n + 1) = m * 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%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%two-rec: ⊦ (m + 2) / 2 = m / 2 + 1; from specification nat-div%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%lez: ⊦ 0 ≤ m; from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%ref-1: ⊦ ¬ n < n; from specification nat-basic1 Used terms: n, m, 0