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Proof step 102 from proof lim-exists (previous step)

ii(0) = 0, # σσ = ∞, mon(σσ), ii(0) = 0 ∧ # σσ = ∞ ∧ mon(σσ) ∧ (∀ 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[n0] = σσ[0] + ⌜nf to ii(n0)⌝ ∨ σσ = [] ∨ (∃ n. # σσ[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 = [], σσ[n0 + 1] = σσ[0] + ⌜nf to ii(n0 + 1)⌝ , ∃ n. # σσ[n] = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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