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Proof step 2 from proof ls-01 (previous step)

m0 * m < m ⊦ m = 0, m0 = 0

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ m0. m0 * m1 < m1 → m1 = 0 ∨ m0 = 0, m0 * m1 +1 < m1 +1 ⊦ m1 +1 = 0, m0 = 0

Rule argument

term: m

Additional information

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