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Proof step 45 from proof infinite-append-prefixnextnli (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 46:
# x = ii(n), ii(n + 1) = ii(n) + # nextnli0(lf(n), n), ii(0) = 0, lf(0) = [], lf(n + 1) = nextnli(lf(n), n), nextnli(x, n) = x + nextnli0(x, n), (x from ii(n)) to # nextnli0(lf(n), n) = nextnli0(lf(n), n), ii(n + 1) ≤ ii(n), ∀ n0. lf(n0) = x to ii(n0), ∀ 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) ⊦ nextnli(x, n) = x

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)