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Proof step 1 from proof pot-mon (conclusion)

1 < m ⊦ m ^ n < m ^ n0 ↔ n < n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ m ^ n < m ^ n0 ↔ n < n0, m = 1, m = 0

Rule argument

Additional information

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