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Proof step 3 from proof monls-mon-implies-monls (previous step)

⊦ mon(nσf) → (∀ n. nσf(n) ⊑ lim(nσf)) ∧ (∀ m. ¬ # lim(nσf) < ⌜m⌝ → (∃ n. lim(nσf) to m ⊑ nσf(n)))

A spec-lemma