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Proof step 54 from proof lim-exists (previous step)

ii(0) = 0, # σσ = ∞, mon(σσ), ∀ n. σσ[n] = σσ[0] + ⌜nf to ii(n)⌝ , ∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. σσ[n + 1] = σσ[n] + ⌜lf(n)⌝ , ∀ n. ii(n + 1) = ii(n) + # lf(n) ∧ (nf from ii(n)) to # lf(n) = lf(n) ⊦ σσ = [], ∃ n. # σσ[n] = ∞, (∃ σ1. σσ[n] + σ1 = σσ[0] + ⌜nf⌝ ) ∧ (∃ n, σ1. (σσ[0] + ⌜nf⌝ to m) + σ1 = σσ[n])

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 3

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