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Proof step 3 from proof numeral-17 (previous step)

m1 < m2, m2 * m + n < m1 * m ⊦ m2 = 0, m1 = 0, m = 0

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
∀ m1, m, n. m1 < m0 ∧ m0 * m + n < m1 * m → m0 = 0 ∨ m1 = 0 ∨ m = 0, m1 < m0 +1, m0 +1 * m + n < m1 * m ⊦ m0 +1 = 0, m1 = 0, m = 0

Rule argument

term: m2

Additional information

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