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Proof step 131 from proof infinite-append-butlast (previous step)

ii(2 + n1) = (m0 + 1) + # x0, ii(0) = 0, ii(n1 + 1) = m0 + 1, lf(2 + n1) = a ' + x1, (σ from m0 + 1) to # x0 = ⌜x0⌝ , x0 = y + a ', a ∈ lf(2 + n1), (finp(σ) ⊃ # σ = ⌜ii(n)⌝ ∧ (∀ m. n ≤ m → (m = 0 ⊃ lf(0);lf(m).tail) = [] ∧ ii(m) = ii(n));∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1)), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. n ≠ 0 → ii(n) ≠ 0 ⊦ σ[m0 + # x0] = a, x0 = [], # σ < ⌜m0⌝ + ⌜1⌝ + ⌜# x0⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 132:
ii(n1 + 1) = m0 + 1, ii(0) = 0, ii(2 + n1) = 2 + m0 + # y, lf(2 + n1) = a ' + x1, (σ from m0 + 1) to # y + 1 = ⌜y⌝ + ⌜a '⌝ , (finp(σ) ⊃ # σ = ⌜ii(n)⌝ ∧ (∀ m. n ≤ m → (m = 0 ⊃ lf(0);lf(m).tail) = [] ∧ ii(m) = ii(n));∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1)), ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ m, n. m ≤ n → ii(m) ≤ ii(n) ⊦ σ[m0 + # y + 1] = a, # σ < ⌜2⌝ + ⌜m0⌝ + ⌜# y⌝

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf% ⊦ ⌜(* n)⌝ + ⌜(* m)⌝ = ⌜n + m⌝ ; (s)% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)% ⊦ x + a ' ≠ []; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ⌜x + y⌝ = ⌜x⌝ + ⌜y⌝ ; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)