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Proof step 117 from proof lim-exists (previous step)

# σσ = ∞, ii(0) = 0, ii(n0 + 1) = ii(n0) + # lf(n0), n < ii(n0 + 1), mon(σσ), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1) ⊦ # σσ[0] = ∞, ((nf to ii(n0)) + ((nf from ii(n0)) to # lf(n0)))[n] = nf(n), ii(n0 + 1) = 0, σσ = [], n < #(nf to ii(n0)), ∃ n. # σσ[n] = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 118:
# σσ = ∞, ii(n0 + 1) = ii(n0) + # lf(n0), ii(0) = 0, n < ii(n0 + 1), ii(n0) < ii(n0 + 1), mon(σσ), ii(n0) ≤ n, ∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1), ∀ m, n. m ≤ n → ii(m) ≤ ii(n) ⊦ # σσ[0] = ∞, ((nf from ii(n0)) to # lf(n0))[n - ii(n0)] = nf(n), ii(n0 + 1) = 0, lf(n0) = [], σσ = [], ∃ n. # σσ[n] = ∞

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ # x = 0 ↔ x = []; (s)% ⊦ m < m + n ↔ n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ #(nf to n) = n; (s)% ⊦ ¬ n < # x → (x + y)[n] = y[n - # x]; (s)