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Proof step 119 from proof infinite-append-butlast (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 120:
lf(n1 + 1) = σ[m0] ' + x0, ii(n1 + 1) = m0 + 1, ii(0) = 0, ii(2 + n1) = (m0 + 1) + # x0, (σ from m0 + 1) to # x0 = ⌜x0⌝ , lf(2 + n1) = a ' + x1, σ[m0] ∈ lf(n1 + 1), ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n).last = lf(n + 1).first ∧ lf(n) ≠ [], (finp(σ) ⊃ # σ = ⌜ii(n)⌝ ∧ (∀ m. n ≤ m → (m = 0 ⊃ lf(0);lf(m).rest) = [] ∧ ii(m) = ii(n));∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1)) ⊦ lf(n1 + 1) = [], σ[m0 + # x0] = a, # σ < ⌜m0⌝ + ⌜1⌝ + ⌜# x0⌝

Rule argument

Additional information

Used simplifier rules: in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%c: ⊦ a + x = a ' + x; from specification list