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Proof step 7 from proof monls-mon-implies-monls (previous step)

mon(nσf), ∀ n. mon<(nσf(n)), ∀ n. nσf(n) ⊑ lim(nσf), ∀ m. ¬ # lim(nσf) < ⌜m⌝ → (∃ n. lim(nσf) to m ⊑ nσf(n)) ⊦ ∀ n. ⌜n⌝ + ⌜1⌝ < # lim(nσf) → lim(nσf)[n] < lim(nσf)[n + 1]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⌜n⌝ + ⌜1⌝ < # lim(nσf), mon(nσf), lim(nσf)[n + 1] ≤ lim(nσf)[n], ∀ n. mon<(nσf(n)), ∀ n. nσf(n) ⊑ lim(nσf), ∀ m. ¬ # lim(nσf) < ⌜m⌝ → (∃ n. lim(nσf) to m ⊑ nσf(n)) ⊦

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat