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Proof step 62 from proof lim-exists (previous step)

# σσ[0] = ⌜n⌝ , 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, σ1. (σσ[0] + ⌜nf⌝ to m) + σ1 = σσ[n], ∃ n. # σσ[n] = ∞

User applied rule cut formula

And obtained the following new goals

Proof step 63:
m < n, # σσ[0] = ⌜n⌝ , 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, σ1. (σσ[0] + ⌜nf⌝ to m) + σ1 = σσ[n], ∃ n. # σσ[n] = ∞
Proof step 66:
# σσ[0] = ⌜n⌝ , 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) ⊦ m < n, σσ = [], ∃ n, σ1. (σσ[0] + ⌜nf⌝ to m) + σ1 = σσ[n], ∃ n. # σσ[n] = ∞

Rule argument

fma: m < n

Additional information

Used formulas: m < n