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Proof step 39 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) ⊦ ∃ n0. ii(n) = ii(n0) ∧ nextnli(x, n0) = x ∧ (∀ m. n0 ≤ m → ii(m) = ii(n0))

User applied rule exists right

And obtained the following new goals

Proof step 40:
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) ⊦ ii(n) = ii(n) ∧ nextnli(x, n) = x ∧ (∀ m. n ≤ m → ii(m) = ii(n))

Rule argument

pos: right 1 terms: n abort: F discard: T computed: T

Additional information

Used terms: n