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Proof step 19 from proof infinite-append-prefixnextnli (previous step)

lf(n0) = (σ to ii(n0)).l, ∀ n0. lf(n0) = (σ to ii(n0)).l, ii(0) = 0, lf(0) = [], ∀ n. ii(n + 1) = ii(n) + # nextnli0(lf(n), n) ∧ ¬ # σ < ⌜ii(n + 1)⌝ ∧ (σ from ii(n)) to # nextnli0(lf(n), n) = ⌜nextnli0(lf(n), n)⌝ , ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n + 1) = nextnli(lf(n), n), ∀ x, n. nextnli(x, n) = x + nextnli0(x, n), (finp(σ) ⊃ # σ = ⌜ii(n)⌝ ∧ (∀ m. n ≤ m → nextnli0(lf(m), m) = [] ∧ ii(m) = ii(n));∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1)) ⊦ σ to ii(n0 + 1) = ⌜nextnli((σ to ii(n0)).l, n0)⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
lf(n0) = (σ to ii(n0)).l, ii(0) = 0, lf(0) = [], ∀ n0. lf(n0) = (σ to ii(n0)).l, ∀ n. ii(n + 1) = ii(n) + # nextnli0(lf(n), n) ∧ ¬ # σ < ⌜ii(n + 1)⌝ ∧ (σ from ii(n)) to # nextnli0(lf(n), n) = ⌜nextnli0(lf(n), n)⌝ , ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n + 1) = nextnli(lf(n), n), ∀ x, n. nextnli(x, n) = x + nextnli0(x, n), (finp(σ) ⊃ # σ = ⌜ii(n)⌝ ∧ (∀ m. n ≤ m → nextnli0(lf(m), m) = [] ∧ ii(m) = ii(n));∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1)) ⊦ σ to ii(n0 + 1) = ⌜nextnli(lf(n0), n0)⌝

Proof step 19 from proof infinite-append-prefixnextnli (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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