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Proof step 42 from proof infinite-append-prefixnextnli (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 43:
ii(0) = 0, lf(0) = [], # x = ii(n), ∀ m. n ≤ m → nextnli0(lf(m), m) = [] ∧ ii(m) = ii(n), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n + 1) = nextnli(lf(n), n), nextnli(x, n) = x + nextnli0(x, n), ∀ n0. lf(n0) = x to ii(n0), ∀ n0. ii(n0 + 1) = ii(n0) + # nextnli0(lf(n0), n0) ∧ ¬ ii(n) < ii(n0 + 1) ∧ (x from ii(n0)) to # nextnli0(lf(n0), n0) = nextnli0(lf(n0), n0) ⊦ nextnli(x, n) = x

Rule argument

pos: left 7 terms: x, n abort: F discard: T computed: T

Additional information

Used terms: x, n