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Proof step 4 from proof numeral-10 (previous step)

m2 < m0 +1, m0 +1 * m + n0 < m2 * m, ∀ m, n0, m2. m0 * m + n0 < m2 * m ∧ m2 < m0 → m = 0 ∨ m2 = 0 ⊦ m = 0, m2 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
(m * m0 + m) + n0 < m2 * m, m2 ≤ m0, ∀ m, n0, m2. m0 * m + n0 < m2 * m ∧ m2 < m0 → m = 0 ∨ m2 = 0 ⊦ m = 0, m2 = 0

Rule argument

Additional information

Used simplifier rules: rsucc-02: ⊦ m * (n + 1) = m * n + m;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)