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Proof step 36 from proof infinite-append-butlast (previous step)

ii(0) = 0, lf(n0) = [], (σ from ii(n0)) to # (n0 = 0 ⊃ lf(0);[].rest) = ⌜(n0 = 0 ⊃ lf(0);[].rest)⌝ , ii(n0 + 1) = ii(n0) + # (n0 = 0 ⊃ lf(0);[].rest), ii(n0 + 1) = ((n0 = 0 ⊃ 0;ii(n0) - 1) - 1) + 1, ii(m) ≤ ii(n1), (n1 = 0 ⊃ 0;ii(n1) - 1) < ii(m) - 1, m ≤ n1, ⌜(n0 = 0 ⊃ 0;ii(n0) - 1) - 1⌝ < # σ, n0 ≠ 0 → 0 = ii(n0) - 1, ∀ 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)), ∀ n. σ[ii(n + 1) - 1] = lf(n + 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 = 0, 0 = ii(m) - 1

Proof step 36 from proof infinite-append-butlast (previous step)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

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