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Proof step 110 from proof infinite-append-butlast (previous step)

σ[0 - 1] = lf(n1 + 1).first, ii(0) = 0, ii(n1 + 1) = 0, σ to ii(2 + n1) = ⌜lf(n1 + 1).rest⌝ , ii(2 + n1) = # lf(n1 + 1).rest, (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)⌝

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 9 terms: n1 + 1

Additional information

Used simplifier rules: oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: n1 + 1