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Proof step 82 from proof lim-exists (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 83:
n1 = n2 + n, # 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), n ≤ n1, ∀ 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) ⊦ ii(m0 + 1) + n = 0, σσ = [], ii(m0 + 1) = 0, ((σσ[0] + ⌜nf⌝ to n0 + n) + ⌜(nf to ii(m0 + 1)) from n0⌝ )[n1] = nf(n2), ∃ n. # σσ[n] = ∞

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n1, n, n2] flags:

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n1, n, n2