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Proof step 104 from proof lim-exists (previous step)

# σσ = ∞, σσ[n0] = σσ[0] + ⌜nf to ii(n0)⌝ , ii(0) = 0, mon(σσ), ∀ n. ii(n + 1) = ii(n) + # lf(n) ∧ (nf from ii(n)) to # lf(n) = lf(n), ∀ n. σσ[n + 1] = σσ[n] + ⌜lf(n)⌝ , ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1) ⊦ σσ[n0 + 1] = σσ[0] + ⌜nf to ii(n0 + 1)⌝ , σσ = [], ∃ n. # σσ[n] = ∞

User applied rule all left

And obtained the following new goals

Proof step 105:
# σσ = ∞, σσ[n0] = σσ[0] + ⌜nf to ii(n0)⌝ , σσ[n0 + 1] = σσ[n0] + ⌜lf(n0)⌝ , ii(0) = 0, mon(σσ), ∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. ii(n + 1) = ii(n) + # lf(n) ∧ (nf from ii(n)) to # lf(n) = lf(n) ⊦ σσ[n0 + 1] = σσ[0] + ⌜nf to ii(n0 + 1)⌝ , σσ = [], ∃ n. # σσ[n] = ∞

Rule argument

pos: left 6 terms: n0 abort: F discard: T computed: T

Additional information

Used terms: n0