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Proof step 98 from proof infinite-append-butlast (previous step)

m0 + 1 = # lf(0), lf(0).last = a, ii(1) = m0 + 1, ii(0) = 0, lf(1) = a + x0, 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. 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 ⊦ σ[m0] = a, lf(0) = [], lf(1) = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 99:
m0 + 1 = # lf(0), ii(0) = 0, ii(1) = m0 + 1, lf(1) = lf(0).last ' + x0, σ to m0 + 1 = ⌜lf(0)⌝ , ⌜m0⌝ < # σ, lf(0).last ∈ lf(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)), ∀ 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(0).last, lf(1) = [], lf(0) = []

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)