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Proof step 6 from proof numeral-10 (previous step)

(m * m0 + m) + n0 < m1 +1 * m, m1 +1 ≤ m0, ∀ m, n0, m2. m0 * m + n0 < m2 * m ∧ m2 < m0 → m = 0 ∨ m2 = 0, ∀ m, m0, n0. (m * m0 + m) + n0 < m1 * m ∧ m1 ≤ m0 ∧ (∀ m, n0, m2. m0 * m + n0 < m2 * m ∧ m2 < m0 → m = 0 ∨ m2 = 0) → m = 0 ∨ m1 = 0 ⊦ m1 +1 = 0, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
m * m0 + n0 < m * m1, m1 < m0, ∀ m, n0, m2. m0 * m + n0 < m2 * m ∧ m2 < m0 → m = 0 ∨ m2 = 0, ∀ m3, m4, n. (m3 * m4 + m3) + n < m1 * m3 ∧ m1 ≤ m4 ∧ (∀ m, n0, m2. m4 * m + n0 < m2 * m ∧ m2 < m4 → m = 0 ∨ m2 = 0) → m3 = 0 ⊦ m = 0, m1 = 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;%zero-06: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0);% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)% ⊦ m +1 = m + 1; (s)