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Proof step 28 from proof infinite-append-butlast (previous step)

(# x0 + m0) + 1 = ii(n0 + 1), m0 + # lf(n0) - 1 = # x0 + m0, ii(0) = 0, ii(n0) = m0 + 1, lf(n0) = a + x0, (σ from m0 + 1) to # x0 = ⌜x0⌝ , ⌜# x0⌝ + ⌜m0⌝ < # σ, ⌜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) ≠ [], ∀ n. σ[ii(n + 1) - 1] = lf(n + 1).head ⊦ lf(n0) = [], ⌜σ[m0] '⌝ + (σ from m0 + 1) to # lf(n0) = ⌜lf(n0)⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
(# x0 + m0) + 1 = ii(n0 + 1), m0 + # lf(n0) - 1 = # x0 + m0, ii(n0) = m0 + 1, ii(0) = 0, lf(n0) = a ' + x0, (σ from m0 + 1) to # x0 = ⌜x0⌝ , ⌜m0⌝ < # σ, ⌜# x0⌝ + ⌜m0⌝ < # σ, a ∈ lf(n0), (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. σ[ii(n + 1) - 1] = lf(n + 1).head, ∀ n. lf(n).last = lf(n + 1).head ∧ lf(n) ≠ [], ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ 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)⌝ , ∀ n. n ≠ 0 → ii(n) ≠ 0 ⊦ lf(n0) = [], ⌜σ[m0] '⌝ + (σ from m0 + 1) to # lf(n0) = ⌜lf(n0)⌝

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list% ⊦ a + x = a ' + x; (s)