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Proof step 1 from proof ls-04 (conclusion)

⊦ m0 ≠ 0 → (m0 < m0 * m ↔ 1 < m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ m0 = 0, m0 < m0 * m ↔ m ≠ 0 ∧ m ≠ 1

Rule argument

Additional information

Used simplifier rules: one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1