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

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

Heuristic structural induction applied rule structural induction

And obtained the following new goals

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

Rule argument

term: m2

Additional information

Used simplifier rules: nat freely generated by 0, +1;;% ⊦ ¬ m < 0; (s)