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Proof step 80 from proof infinite-append-butlast (previous step)

lf(n0) = a + x0 ∧ lf(n0) ≠ [], (σ from m2 + 1) to # x0 = ⌜x0⌝ , ii(0) = 0, ii(n0 + 1) = (m2 + # x0) + 1, m2 + # lf(n0) = m0 + 1, ii(n0) = m2 + 1, ⌜m2⌝ + ⌜# x0⌝ < # σ, (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)), ∀ n. lf(n).last = lf(n + 1).first ∧ lf(n) ≠ [], ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. ii(n + 1) = ii(n) + # (n = 0 ⊃ lf(0);lf(n).rest) ∧ ¬ # σ < ⌜ii(n + 1)⌝ ∧ (σ from ii(n)) to # (n = 0 ⊃ lf(0);lf(n).rest) = ⌜(n = 0 ⊃ lf(0);lf(n).rest)⌝ , ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ n. σ[ii(n + 1) - 1] = lf(n + 1).first ⊦ m2 + # x0 = m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 81:
(σ from m2 + 1) to # x0 = ⌜x0⌝ , ii(0) = 0, ii(n0 + 1) = (m2 + # x0) + 1, m2 + # lf(n0) = m0 + 1, ii(n0) = m2 + 1, lf(n0) = a ' + x0, ⌜m2⌝ + ⌜# x0⌝ < # σ, a ∈ lf(n0), (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)), ∀ n. lf(n).last = lf(n + 1).first ∧ lf(n) ≠ [], ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. ii(n + 1) = ii(n) + # (n = 0 ⊃ lf(0);lf(n).rest) ∧ ¬ # σ < ⌜ii(n + 1)⌝ ∧ (σ from ii(n)) to # (n = 0 ⊃ lf(0);lf(n).rest) = ⌜(n = 0 ⊃ lf(0);lf(n).rest)⌝ , ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ n. σ[ii(n + 1) - 1] = lf(n + 1).first ⊦ m2 + # x0 = m0, 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%c: ⊦ a + x = a ' + x; from specification list