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Proof step 116 from proof infinite-append-butlast (previous step)

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] = lf(2 + n1).first, # σ < ⌜m0⌝ + ⌜1⌝ + ⌜# x0⌝

User applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: list-data inst: name: elim-1 seq: ⊦ x ≠ [] → (a = x.first ∧ x0 = x.rest ↔ x = a + x0) subst: [x, a, x0] → [lf(2 + n1), a, x] flags: []

Additional information

Used simplifier rules: ⊦ x ≠ [] → (a = x.first ∧ x0 = x.rest ↔ x = a + x0); (elim) Used terms: lf(2 + n1), a, x