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Proof step 115 from proof infinite-append-butlast (previous step)

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

User applied rule insert elim lemma

And obtained the following new goals

Proof step 116:
σ[m0] = lf(n1 + 1).head, ii(n1 + 1) = m0 + 1, ii(0) = 0, ii(2 + n1) = (m0 + 1) + # lf(n1 + 1).tail, lf(n1 + 1) = [], (σ from m0 + 1) to # lf(n1 + 1).tail = ⌜lf(n1 + 1).tail⌝ , (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, # σ < ⌜ii(2 + n1)⌝
Proof step 118:
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)⌝

Rule argument

spec: list-data name: elim-1 seq: ⊦ x ≠ [] → (a = x.head ∧ x0 = x.tail ↔ x = a + x0) subst: [x, a, x0] → [lf(n1 + 1), a, x]

Additional information

Used simplifier rules: ⊦ x ≠ [] → (a = x.head ∧ x0 = x.tail ↔ x = a + x0); (elim) Used terms: lf(n1 + 1), a, x