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Proof step 8 from proof mult-le-01 (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 9:
m ≤ n, n * n ≤ m + 1 ⊦ m * n ≤ m + 1, n ≤ 1 ∧ m ≤ 1

Rule argument

pos: right 1

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;% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≠ 0 → (m ≤ 1 ↔ m = 1); (s)