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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: <toplevel> 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)]

Additional information

Used simplifier rules: max-zero: ⊦ maxenc(m) = 0 ↔ m = 0;% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m * (n + 1) = m * n + m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ (m + 2) / 2 = m / 2 + 1; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)%m0 ≠ 0 ⊦ n ≤ m / m0 ↔ n * m0 ≤ m; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ (m + n) / 2 < n ↔ m < n; (s)% ⊦ m0 * m < m ↔ m0 = 0 ∧ m ≠ 0; (ws) Used terms: m, maxenc(m)