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Proof step 3 from proof modls (previous step)

m0 * m + n0 < m0 * n, n0 < m0 ⊦ m < n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n0 < m0, m0 * m + n0 < m0 * n ⊦ n = 0, m < n

Rule argument

Additional information

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