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

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

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 3

Additional information

Used simplifier rules: mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%sqr-08: ⊦ m * m ≤ 2 ↔ m ≤ 1;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)%not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward) from specification nat%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m ≠ 1 → (m < 2 ↔ m = 0); (s)