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Proof step 1 from proof monls-mon-implies-monls (conclusion)

⊦ (∀ n. mon<(nσf(n))) ∧ mon(nσf) → (¬ mon<(lim(λ n. nσf(n))) ↔ false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information