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Proof step 118 from proof infinite-append-butlast (previous step)

σ[m0] = a, ii(n1 + 1) = m0 + 1, ii(0) = 0, ii(2 + n1) = (m0 + 1) + # x0, lf(n1 + 1) = a + x0, (σ from m0 + 1) to # x0 = ⌜x0⌝ , (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), ∀ n. lf(n).last = lf(n + 1).head ∧ lf(n) ≠ [] ⊦ σ[ii(2 + n1) - 1] = lf(2 + n1).head, lf(n1 + 1) = [], # σ < ⌜ii(2 + n1)⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 119:
ii(n1 + 1) = m0 + 1, ii(0) = 0, ii(2 + n1) = (m0 + 1) + # x0, lf(n1 + 1) = σ[m0] ' + x0, (σ from m0 + 1) to # x0 = ⌜x0⌝ , σ[m0] ∈ lf(n1 + 1), (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), ∀ n. lf(n).last = lf(n + 1).head ∧ lf(n) ≠ [] ⊦ σ[m0 + # x0] = lf(2 + n1).head, lf(n1 + 1) = [], # σ < ⌜m0⌝ + ⌜1⌝ + ⌜# x0⌝

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list% ⊦ a + x = a ' + x; (s)