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Proof step 15 from proof infinite-append-prefixnextnli (previous step)

ii(0) = 0, lf(0) = [], ii(n0 + 1) < ii(n0), ∀ 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), (finp(σ) ⊃ # σ = ⌜ii(n)⌝ ∧ (∀ m. n ≤ m → nextnli0(lf(m), m) = [] ∧ ii(m) = ii(n));∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1)) ⊦ ii(n0) = 0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: n0, n0 + 1

Additional information

Used simplifier rules: lep: ⊦ m ≤ m + n; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat Used terms: n0, n0 + 1