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Proof step 17 from proof monls-mon-implies-monls (previous step)

(lim(nσf) to n + 2) + nσ = nσf(n0), ⌜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⌝

User applied rule all left

And obtained the following new goals

Proof step 18:
(lim(nσf) to n + 2) + nσ = nσf(n0), ⌜n⌝ + ⌜1⌝ < # lim(nσf), mon(nσf), lim(nσf)[n + 1] ≤ lim(nσf)[n], mon<(nσf(n0)), ∀ n. nσf(n) ⊑ lim(nσf) ⊦ # lim(nσf) < ⌜n⌝ + ⌜2⌝

Rule argument

pos: left 5 terms: n0 abort: F discard: T computed: F

Additional information

Used terms: n0