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Proof step 113 from proof infinite-append-butlast (previous step)

(σ from m0 + 1) to # lf(n1 + 1).rest = ⌜lf(n1 + 1).rest⌝ , ii(2 + n1) = (m0 + 1) + # lf(n1 + 1).rest, σ[m0] = lf(n1 + 1).first, ii(0) = 0, ii(n1 + 1) = m0 + 1, (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. n ≠ 0 → ii(n) ≠ 0 ⊦ lf(n1 + 1) ≠ [], σ[ii(2 + n1) - 1] = lf(2 + n1).first, # σ < ⌜ii(2 + n1)⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 114:
lf(n1 + 1) = [], (σ from m0 + 1) to # [].rest = ⌜[].rest⌝ , ii(2 + n1) = (m0 + 1) + # [].rest, σ[m0] = [].first, ii(0) = 0, ii(n1 + 1) = m0 + 1, (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. n ≠ 0 → ii(n) ≠ 0 ⊦ σ[ii(2 + n1) - 1] = lf(2 + n1).first, # σ < ⌜ii(2 + n1)⌝

Proof step 113 from proof infinite-append-butlast (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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