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Proof step 5 from proof max-square (previous step)

⊦ ((m + n) * (m + n + 1)) / 2 + n < (((m + n) + 1) * ((m + n) + 2)) / 2, maxenc(((m + n) * (m + n + 1)) / 2 + n) = m + n

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 6:
⊦ m0 ≤ m ∧ m ≤ n → (m * (m + 1)) / 2 + m0 < ((n + 1) * (n + 2)) / 2

Rule argument

spec: nat-div inst: name: square-ls-03 seq: ⊦ m0 ≤ m ∧ m ≤ n → (m * (m + 1)) / 2 + m0 < ((n + 1) * (n + 2)) / 2 subst: [m0, m, n] → [n, m + n, m + n] flags:

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%lep: ⊦ m ≤ m + n; from specification nat%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%ler: ⊦ m ≤ m; from specification nat Used terms: n, m + n, m + n