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Proof step 5 from proof mult-le-01 (previous step)

m = 0 ∨ m = 1 ∧ n ≤ 2 ∨ 2 ≤ m ∧ n < 2, m ≤ n ⊦ n * n ≤ m + 1 ↔ n ≤ 1 ∧ m ≤ 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m ≤ n, n * n ≤ m + 1, 2 ≤ m ∧ n < 2 ∨ m = 1 ∧ n ≤ 2 ⊦ m = 0, 1 = n ∧ m = 1

Rule argument

Additional information

Used simplifier rules: mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%sqle-02: ⊦ m * m ≤ n ∧ n ≤ n0 → m ≤ n0;%sqzero: ⊦ m * m = 0 ↔ m = 0;%one-01: ⊦ m0 * m = 1 ↔ m0 = 1 ∧ m = 1;%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n); from specification nat%one-02: ⊦ m ≠ 0 → (m ≤ 1 ↔ m = 1); from specification nat%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%lez: ⊦ 0 ≤ m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2