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Proof step 34 from proof infinite-append-prefixnextnli (previous step)

#(x to ii(n0 + 1)) = # lf(n0 + 1), lf(n0 + 1) = lf(n0) + nextnli0(lf(n0), n0), lf(0) = [], ii(0) = 0, lf(n0) = x to ii(n0), # x = ii(n), n1 < # lf(n0 + 1), ∀ n2. ii(n2 + 1) = ii(n2) + # nextnli0(lf(n2), n2) ∧ ¬ ii(n) < ii(n2 + 1) ∧ (x from ii(n2)) to # nextnli0(lf(n2), n2) = nextnli0(lf(n2), n2), ∀ n2. lf(n2) = x to ii(n2), ∀ n2. n2 < # lf(n0 + 1) → (x to ii(n0 + 1))[n2] = lf(n0 + 1)[n2], ∀ x, n. nextnli(x, n) = x + nextnli0(x, n), ∀ n. lf(n + 1) = nextnli(lf(n), n), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ m. n ≤ m → nextnli0(lf(m), m) = [] ∧ ii(m) = ii(n) ⊦ (x to ii(n0 + 1))[n1] = lf(n0 + 1)[n1], lf(n0 + 1) = []

Proof step 34 from proof infinite-append-prefixnextnli (previous step)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

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