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Proof step 45 from proof lim-exists (previous step)

# σσ[n] = ⌜n0⌝ , # x = ii(n), # σσ = ∞, ii(0) = 0, mon(σσ), n ≤ m0 + 1, # σσ = ∞ ∧ # x = ii(n) ∧ # σσ[n] = ⌜n0⌝ ∧ ii(0) = 0 ∧ mon(σσ) ∧ n ≤ m0 ∧ (∀ m. n ≤ m → lf(m) = [] ∧ ii(m) = ii(n)) ∧ (∀ m, n. m ≤ n → ii(m) ≤ ii(n)) ∧ (∀ n. σσ[n + 1] = σσ[n] + ⌜lf(n)⌝ ) ∧ (∀ n0. ii(n0 + 1) = ii(n0) + # lf(n0) ∧ ¬ ii(n) < ii(n0 + 1) ∧ (x from ii(n0)) to # lf(n0) = lf(n0)) → σσ[m0] = σσ[n] ∨ σσ = [] ∨ (∃ n. # σσ[n] = ∞), ∀ 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) ⊦ σσ[m0 + 1] = σσ[n], σσ = [], ∃ n. # σσ[n] = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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