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Proof step 3 from proof square-ls-03 (previous step)

((n + 1) * (n + 2)) / 2 < ((m + 1) * (m + 2)) / 2 ↔ n < m ⊦ m0 ≤ m ∧ m ≤ n → (m * (m + 1)) / 2 + m0 < ((n + 1) * (n + 2)) / 2

User applied rule case distinction

And obtained the following new goals

Proof step 4:
⊦ ((n + 1) * (n + 2)) / 2 < ((m + 1) * (m + 2)) / 2, n < m, m0 ≤ m ∧ m ≤ n → (m * (m + 1)) / 2 + m0 < ((n + 1) * (n + 2)) / 2

Rule argument

pos: left 1

Additional information

Used simplifier rules: two-rec: ⊦ (m + 2) / 2 = m / 2 + 1;%ls-02: n ≠ 0 ⊦ m / n < m0 / n → m < m0;%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%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%ref-1: ⊦ ¬ n < n; from specification nat-basic1