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Proof step 69 from proof lim-exists (previous step)

# σσ = ∞, # σσ[0] = ⌜n⌝ , ii(0) = 0, mon(σσ), n ≤ m, ∀ n. σσ[n] = σσ[0] + ⌜nf to ii(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), ∃ m0. ii(m0) ≤ m - n ∧ m - n < ii(m0 + 1) ⊦ σσ = [], ∃ n. # σσ[n] = ∞, ∃ n, σ1. (σσ[0] + ⌜nf⌝ to m) + σ1 = σσ[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 70:
# σσ = ∞, # σσ[0] = ⌜n⌝ , ii(0) = 0, m < ii(m0 + 1) + n, ii(m0) < ii(m0 + 1), mon(σσ), ii(m0) + n ≤ m, n ≤ m, ∀ n. σσ[n] = σσ[0] + ⌜nf to ii(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) + n = 0, ii(m0 + 1) = 0, σσ = [], ∃ n. # σσ[n] = ∞, ∃ n, σ1. (σσ[0] + ⌜nf⌝ to m) + σ1 = σσ[n]

Rule argument

Additional information

Used simplifier rules: ⊦ m < m + n ↔ n ≠ 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; (s)%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)