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Proof step 7 from proof mult-le-01 (previous step)

m ≤ n ⊦ m * n ≤ m + 1, n * n ≤ m + 1 ↔ n ≤ 1 ∧ m ≤ 1

User applied rule case distinction

And obtained the following new goals

Proof step 8:
n * n ≤ m + 1, m ≤ n ⊦ n ≤ 1 ∧ m ≤ 1, m * n ≤ m + 1

Rule argument

pos: right 2

Additional information

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