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Proof step 45 from proof infinite-append-prefixnextli-weak (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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