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Proof step 1 from proof ls-08 (conclusion)

⊦ m * n + n0 < n ↔ m = 0 ∧ n0 < n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
m * n + n0 < n, n0 < n ⊦ n = 0, m = 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;%nls-fst: ⊦ ¬ m < n → ¬ m + m0 < n; from specification nat-basic2%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2