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Proof step 6 from proof numeral-17 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
m2 < m0, m0 * m + n < m2 * m, ∀ m1, m3, n0. m1 < m0 ∧ m0 * m3 + n0 < m1 * m3 → m1 = 0 ∨ m3 = 0, ∀ m3, m4, n0. (m3 * m4 + m4) + n0 < m2 * m4 ∧ m2 ≤ m3 ∧ (∀ m1, m, n. m1 < m3 ∧ m3 * m + n < m1 * m → m3 = 0 ∨ m1 = 0 ∨ m = 0) → m4 = 0 ⊦ m = 0, m0 = 0, m2 = 0

Rule argument

Additional information

Used simplifier rules: rsucc-01: ⊦ (m + 1) * n = m * n + n;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%zero-06: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0);%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2