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Proof step 105 from proof infinite-append-butlast (previous step)

ii(0) = 0, ii(0) = 0 ∧ (∀ n. n ≠ 0 → ii(n) ≠ 0) ∧ (∀ 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)⌝ ) ∧ (∀ m, n. m ≤ n → ii(m) ≤ ii(n)) ∧ (∀ n. lf(n).last = lf(n + 1).first ∧ lf(n) ≠ []) ∧ (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)) → σ[ii(n1 + 1) - 1] = lf(n1 + 1).first, ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ 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)⌝ , ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n).last = lf(n + 1).first ∧ lf(n) ≠ [], (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)) ⊦ σ[ii(2 + n1) - 1] = lf(2 + n1).first

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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