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Proof step 74 from proof lim-exists (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 75:
# σσ = ∞, ii(0) = 0, # σσ[0] = ⌜n⌝ , σσ[m0] = σσ[0] + ⌜nf to ii(m0)⌝ , σσ[m0 + 1] = σσ[0] + ⌜nf to ii(m0 + 1)⌝ , mon(σσ), n0 < ii(m0 + 1), ii(m0) ≤ n0, ii(m0) < 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, (σσ[0] + ⌜nf⌝ to n0 + n) + ⌜(nf to ii(m0 + 1)) from n0⌝ = σσ[m0 + 1], ∃ n. # σσ[n] = ∞

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%lep: ⊦ m ≤ m + n; from specification nat%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2