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Proof step 149 from proof infinite-append-prefixnextli (previous step)

# x = ii(n), ii(m0 + 1) = ii(m0) + # nextli0(lf(m0)), ii(0) = 0, lf(m0 + 1) = lf(m0) + nextli0(lf(m0)), lf(0) = [], lf(2 + m0) = nextli(lf(m0) + nextli0(lf(m0))), nextli0(lf(n0)) = [], (x from ii(m0)) to # nextli0(lf(m0)) = nextli0(lf(m0)), n0 ≤ m0, ii(m0 + 1) ≤ ii(n), n0 < m0 → lf(m0) = lf(n0) ∧ nextli0(lf(m0)) = [] ∧ ii(m0) = ii(n0), ∀ 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), ∀ n0. lf(n0) = x to ii(n0), ∀ n0. ii(n0 + 1) = ii(n0) + # nextli0(lf(n0)) ∧ ¬ ii(n) < ii(n0 + 1) ∧ (x from ii(n0)) to # nextli0(lf(n0)) = nextli0(lf(n0)) ⊦ lf(m0) + nextli0(lf(m0)) = lf(n0) ∧ nextli0(lf(m0) + nextli0(lf(m0))) = [] ∧ ii(m0 + 1) = ii(n0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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