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Proof step 14 from proof monls-mon-implies-monls (previous step)

⌜n⌝ + ⌜1⌝ < # lim(nσf), mon(nσf), lim(nσf)[n + 1] ≤ lim(nσf)[n], lim(nσf) to n + 2 ⊑ nσf(n0), ∀ n. mon<(nσf(n)), ∀ n. nσf(n) ⊑ lim(nσf) ⊦ # lim(nσf) < ⌜n⌝ + ⌜2⌝

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ nσ ⊑ nσ0 ↔ (∃ nσ1. nσ + nσ1 = nσ0)
Proof step 16:
⌜n⌝ + ⌜1⌝ < # lim(nσf), mon(nσf), lim(nσf)[n + 1] ≤ lim(nσf)[n], ∃ nσ1. (lim(nσf) to n + 2) + nσ1 = nσf(n0), ∀ n. mon<(nσf(n)), ∀ n. nσf(n) ⊑ lim(nσf) ⊦ # lim(nσf) < ⌜n⌝ + ⌜2⌝

Rule argument

spec: mon-lim-filter-count inst: natstream name: prefix-def seq: ⊦ nσ ⊑ nσ0 ↔ (∃ nσ1. nσ + nσ1 = nσ0) subst: [nσ0, nσ] → [nσf(n0), lim(nσf) to n + 2] flags: rewrite

Additional information

Used terms: nσf(n0), lim(nσf) to n + 2