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Proof step 93 from proof infinite-append-prefixnextli (previous step)

nextli0(lf(n)) = [], ii(0) = 0, lf(0) = [], # lf(n) = ii(n), nextli(lf(n)) = lf(n), n0 < n, ∀ n0. lf(n0) = lf(n) to ii(n0), ∀ n0. ii(n0 + 1) = ii(n0) + # nextli0(lf(n0)) ∧ ¬ ii(n) < ii(n0 + 1) ∧ (lf(n) from ii(n0)) to # nextli0(lf(n0)) = nextli0(lf(n0)), nextli0(lf(n0)) = [] ↔ n0 ≤ n0, ∀ m. nextli0(lf(m)) = [] ↔ n0 ≤ m, ∀ n. nextli0(lf(n)) = [] → (∀ m. n < m → lf(m) = lf(n) ∧ nextli0(lf(m)) = [] ∧ ii(m) = ii(n)), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n + 1) = nextli(lf(n)), ∀ x. nextli(x) = x + nextli0(x), ∀ m. n ≤ m → nextli0(lf(m)) = [] ∧ ii(m) = ii(n) ⊦ n = 0, ii(n) = ii(n0) ∧ (n0 ≤ m ⊃ ii(m) = ii(n0);ii(m) < ii(m + 1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 94:
nextli0(lf(n)) = [], ii(0) = 0, lf(0) = [], # lf(n) = ii(n), nextli(lf(n)) = lf(n), nextli0(lf(n0)) = [], n0 < n, ∀ n0. lf(n0) = lf(n) to ii(n0), ∀ n0. ii(n0 + 1) = ii(n0) + # nextli0(lf(n0)) ∧ ¬ ii(n) < ii(n0 + 1) ∧ (lf(n) from ii(n0)) to # nextli0(lf(n0)) = nextli0(lf(n0)), ∀ m. nextli0(lf(m)) = [] ↔ n0 ≤ m, ∀ n. nextli0(lf(n)) = [] → (∀ m. n < m → lf(m) = lf(n) ∧ nextli0(lf(m)) = [] ∧ ii(m) = ii(n)), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n + 1) = nextli(lf(n)), ∀ x. nextli(x) = x + nextli0(x), ∀ m. n ≤ m → nextli0(lf(m)) = [] ∧ ii(m) = ii(n) ⊦ n = 0, ii(n) = ii(n0) ∧ (n0 ≤ m ⊃ ii(m) = ii(n0);ii(m) < ii(m + 1))

Rule argument

Additional information

Used simplifier rules: ler: ⊦ m ≤ m; from specification nat