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Proof step 91 from proof lim-exists (previous step)

⌜n1⌝ < # σσ[0], σσ[m0] = σσ[0] + ⌜nf to ii(m0)⌝ , # σσ[0] = ⌜n⌝ , ii(0) = 0, # σσ = ∞, ii(m0) < ii(m0 + 1), ii(m0) ≤ n0, n0 < ii(m0 + 1), mon(σσ), n1 < ii(m0 + 1) + 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) ⊦ ii(m0 + 1) = 0, σσ = [], ii(m0 + 1) + n = 0, ((σσ[0] + ⌜nf⌝ to n0 + n) + ⌜(nf to ii(m0 + 1)) from n0⌝ )[n1] = σσ[0][n1], ∃ n. # σσ[n] = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 92:
# σσ = ∞, ii(0) = 0, # σσ[0] = ⌜n⌝ , σσ[m0] = σσ[0] + ⌜nf to ii(m0)⌝ , n1 < ii(m0 + 1) + n, mon(σσ), n0 < ii(m0 + 1), ii(m0) ≤ n0, ii(m0) < ii(m0 + 1), n1 < n, ∀ 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) ⊦ ((σσ[0] + ⌜nf⌝ to n0 + n) + ⌜(nf to ii(m0 + 1)) from n0⌝ )[n1] = σσ[0][n1], ii(m0 + 1) + n = 0, σσ = [], ii(m0 + 1) = 0, n = 0, ∃ n. # σσ[n] = ∞

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf