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Proof step 1 from proof max-le (conclusion)

⊦ maxenc(m) ≤ m

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
⊦ maxenc(m) = n → (n * (n + 1)) / 2 ≤ m ∧ m < ((n + 1) * (n + 2)) / 2

Rule argument

spec: nat-pair inst: name: max-rev-01 seq: ⊦ maxenc(m) = n → (n * (n + 1)) / 2 ≤ m ∧ m < ((n + 1) * (n + 2)) / 2 subst: [m, n] → [m, maxenc(m)] flags: current, []

Additional information

Used simplifier rules: max-zero: ⊦ maxenc(m) = 0 ↔ m = 0;%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%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%le-div-to-mult: m0 ≠ 0 ⊦ n ≤ m / m0 ↔ n * m0 ≤ m; from specification nat-div%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (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 Used terms: m, maxenc(m)