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Proof step 6 from proof le (previous step)

m ≤ n ⊦ m0 * m ≤ m0 * n, m0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
m0 * n < m0 * m, m ≤ n ⊦ m0 = 0, m = 0

Rule argument

Additional information

Used simplifier rules: zero-06: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0);%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat