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Proof step 8 from proof monls-mon-implies-monls (previous step)

⌜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)) ⊦

User applied rule all left

And obtained the following new goals

Proof step 9:
⌜n⌝ + ⌜1⌝ < # lim(nσf), mon(nσf), lim(nσf)[n + 1] ≤ lim(nσf)[n], ∀ n. mon<(nσf(n)), ∀ n. nσf(n) ⊑ lim(nσf), ¬ # lim(nσf) < ⌜n + 2⌝ → (∃ n0. lim(nσf) to n + 2 ⊑ nσf(n0)) ⊦

Rule argument

pos: left 6 terms: n + 2 abort: F discard: T computed: F

Additional information

Used terms: n + 2