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

ii(0) = 0, lf(0) = [], lf(n0 + 1) = lf(n0) + nextnli0(lf(n0), n0), ii(n0 + 1) = # lf(n0 + 1), lf(n0) = nf to ii(n0), n1 < ii(n0 + 1), ∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1), ∀ 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), ∀ n. lf(n) = nf to ii(n), ∀ n. n < ii(n0 + 1) → nf(n) = lf(n0 + 1)[n], ∀ n. ii(n + 1) = ii(n) + # nextnli0(lf(n), n) ∧ (nf from ii(n)) to # nextnli0(lf(n), n) = nextnli0(lf(n), n) ⊦ ii(n0 + 1) = 0, nf(n1) = lf(n0 + 1)[n1]

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 12 terms: n1

Additional information

Used terms: n1