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Proof step 17 from proof infinite-append-prefixnextnli (previous step)

# σ = ⌜ii(n)⌝ , ii(0) = 0, lf(0) = [], ii(n) < ii(n0 + 1), ∀ n0. lf(n0) = (σ to ii(n0)).l, ∀ n. ii(n + 1) = ii(n) + # nextnli0(lf(n), n) ∧ ¬ # σ < ⌜ii(n + 1)⌝ ∧ (σ from ii(n)) to # nextnli0(lf(n), n) = ⌜nextnli0(lf(n), n)⌝ , ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n + 1) = nextnli(lf(n), n), ∀ x, n. nextnli(x, n) = x + nextnli0(x, n), ∀ m. n ≤ m → nextnli0(lf(m), m) = [] ∧ ii(m) = ii(n) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: n0

Additional information

Used simplifier rules: lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: n0