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Proof step 98 from proof infinite-append-prefixnextli (previous step)

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

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

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