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Proof step 95 from proof infinite-append-butlast (previous step)

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

User applied rule insert elim lemma

And obtained the following new goals

Rule argument

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

Additional information

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