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Proof step 64 from proof infinite-append-prefixnextli-weak (previous step)

# lf(n0) = ii(n0), ii(0) = 0, lf(0) = [], (σ from ii(n0)) to # nextli0(lf(n0)) = ⌜nextli0(lf(n0))⌝ , n1 < ii(n0) + # nextli0(lf(n0)), (finp(σ) ⊃ # σ = ⌜ii(n)⌝ ∧ (∀ m. n ≤ m → nextli0(lf(m)) = [] ∧ ii(m) = ii(n));∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1)), ∀ n2. n2 < ii(n0) → lf(n0)[n2] = σ[n2], ∀ x. nextli(x) = x + nextli0(x), ∀ n. lf(n + 1) = nextli(lf(n)), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. ii(n + 1) = ii(n) + # nextli0(lf(n)) ∧ ¬ # σ < ⌜ii(n + 1)⌝ ∧ (σ from ii(n)) to # nextli0(lf(n)) = ⌜nextli0(lf(n))⌝ ⊦ (lf(n0) + nextli0(lf(n0)))[n1] = σ[n1], # σ < ⌜ii(n0)⌝ + ⌜# nextli0(lf(n0))⌝ , n1 < # lf(n0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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)%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% ⊦ ¬ n < # x → (x + y)[n] = y[n - # x]; (s)