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Proof step 90 from proof lim-exists (previous step)

# σσ = ∞, ii(0) = 0, # σσ[0] = ⌜n⌝ , σσ[m0] = σσ[0] + ⌜nf to ii(m0)⌝ , mon(σσ), ii(m0) ≤ n0, ii(m0) < ii(m0 + 1), n2 < ii(m0 + 1), n0 ≤ n2, ii(m0) ≤ n2, n0 < ii(m0 + 1), ∀ 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, nf(n0 + (n2 - n0)) = nf(n2), ∃ n. # σσ[n] = ∞

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n2, n0, n1