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Proof step 6 from proof le-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
m + 1 < m * n, 2 ≤ m ∧ n = 1 ∨ m = 1 ∧ n ≤ 2 ⊦ m = 0, n = 0

Rule argument

Additional information

Used simplifier rules: zero: ⊦ 0 * n = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%zero-06: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0);% ⊦ m ≠ 0 → (m < 2 ↔ m = 1); (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)