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Proof step 5 from proof numeral-17 (previous step)

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

User applied rule structural induction

And obtained the following new goals

Proof step 6:
∀ 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

Rule argument

indvar: m1

Additional information

Used simplifier rules: nat freely generated by 0, +1;;