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Proof step 127 from proof infinite-append-butlast (previous step)

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

User applied rule insert elim lemma

And obtained the following new goals

Proof step 128:
x0 = y + a ', 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, a ∈ lf(2 + n1), ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ m, n. m ≤ n → ii(m) ≤ ii(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)) ⊦ σ[m0 + # x0] = a, x0 = [], # σ < ⌜m0⌝ + ⌜1⌝ + ⌜# x0⌝

Rule argument

spec: list-last inst: name: last-elim seq: x ≠ [] ⊦ a = x.last ∧ y = x.butlast ↔ x = y + a ' subst: [x, a, y] → [x0, a, y] flags: []

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.last ∧ y = x.butlast ↔ x = y + a '; (elim) Used terms: x0, a, y