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Proof step 124 from proof infinite-append-butlast (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 125:
ii(n1 + 1) = m0 + 1, ii(0) = 0, ii(2 + n1) = (m0 + 1) + # x0, lf(n1 + 1) = σ[m0] ' + x0, lf(2 + n1) = lf(n1 + 1).last ' + x1, (σ from m0 + 1) to # x0 = ⌜x0⌝ , σ[m0] ∈ lf(n1 + 1), lf(n1 + 1).last ∈ lf(2 + n1), (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) ⊦ σ[m0 + # x0] = lf(n1 + 1).last, lf(n1 + 1) = [], # σ < ⌜m0⌝ + ⌜1⌝ + ⌜# x0⌝

Rule argument

Additional information

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