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

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

User applied rule all left

And obtained the following new goals

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

Rule argument

pos: left 8 terms: n0 abort: F discard: T computed: F

Additional information

Used terms: n0