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Proof step 100 from proof lim-exists (previous step)

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

User applied rule structural induction

And obtained the following new goals

Proof step 101:
# σσ = ∞, ii(0) = 0, 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, ii, lf, nf. # σσ = ∞ ∧ ii(0) = 0 ∧ mon(σσ) ∧ (∀ 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)) → σσ[n0] = σσ[0] + ⌜nf to ii(n0)⌝ ∨ σσ = [] ∨ (∃ n. # σσ[n] = ∞) ⊦ σσ[n0 +1] = σσ[0] + ⌜nf to ii(n0 +1)⌝ , σσ = [], ∃ n. # σσ[n] = ∞

Rule argument

indvar: n

Additional information

Used simplifier rules: nat freely generated by 0, +1;;% ⊦ σ0 + σ = σ0 ↔ σ = [] ∨ ¬ finp(σ0); (s)% ⊦ ⌜x⌝ = [] ↔ x = []; (s)% ⊦ nf to n = [] ↔ n = 0; (s)%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream