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Proof step 68 from proof lim-exists (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 69:
# σσ = ∞, # σσ[0] = ⌜n⌝ , ii(0) = 0, mon(σσ), n ≤ m, ∀ n. σσ[n] = σσ[0] + ⌜nf to ii(n)⌝ , ∀ 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), ∃ m0. ii(m0) ≤ m - n ∧ m - n < ii(m0 + 1) ⊦ σσ = [], ∃ n. # σσ[n] = ∞, ∃ n, σ1. (σσ[0] + ⌜nf⌝ to m) + σ1 = σσ[n]

Rule argument

pos: left 9 terms: m - n abort: F discard: T computed: F

Additional information

Used terms: m - n