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Proof step 34 from proof lim-exists (previous step)

# x = ii(n), # σσ = ∞, ii(0) = 0, # σσ[n] = ⌜n0⌝ , mon(σσ), ∀ n0. ii(n0 + 1) = ii(n0) + # lf(n0) ∧ ¬ ii(n) < ii(n0 + 1) ∧ (x from ii(n0)) to # lf(n0) = lf(n0), ∀ n. σσ[n + 1] = σσ[n] + ⌜lf(n)⌝ , ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ m. n ≤ m → lf(m) = [] ∧ ii(m) = ii(n), ∀ m. n ≤ m → σσ[m] = σσ[n] ⊦ ∃ σ1. σσ[n1] + σ1 = σσ[n], σσ = [], ∃ n. # σσ[n] = ∞

User applied rule all left

And obtained the following new goals

Proof step 35:
# x = ii(n), # σσ = ∞, ii(0) = 0, # σσ[n] = ⌜n0⌝ , mon(σσ), ∀ n0. ii(n0 + 1) = ii(n0) + # lf(n0) ∧ ¬ ii(n) < ii(n0 + 1) ∧ (x from ii(n0)) to # lf(n0) = lf(n0), ∀ n. σσ[n + 1] = σσ[n] + ⌜lf(n)⌝ , ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ m. n ≤ m → lf(m) = [] ∧ ii(m) = ii(n), n ≤ n1 → σσ[n1] = σσ[n], ∀ m. n ≤ m → σσ[m] = σσ[n] ⊦ ∃ σ1. σσ[n1] + σ1 = σσ[n], σσ = [], ∃ n. # σσ[n] = ∞

Rule argument

pos: left 10 terms: n1 abort: F discard: F computed: F

Additional information

Used terms: n1