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Proof step 79 from proof infinite-append-butlast (previous step)

ii(n0) = 2 + m0, lf(n0) = [], (σ from 2 + m0) to # [].rest = ⌜[].rest⌝ , ii(0) = 0, ii(n0 + 1) = 2 + m0 + # [].rest, (⌜m0⌝ + ⌜1⌝ ) + ⌜# [].rest⌝ < # 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 ⊦

Proof step 79 from proof infinite-append-butlast (previous step)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

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