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Proof step 80 from proof infinite-append-prefixnextnli (previous step)

#((σ from ii(n0)) to # nextnli0(lf(n0), n0)) = ⌜# nextnli0(lf(n0), n0)⌝ , ii(0) = 0, lf(0) = [], # lf(n0) = ii(n0), n1 < ii(n0) + # nextnli0(lf(n0), n0), ∀ m, n. m ≤ n → ii(m) ≤ ii(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)), ∀ n2. n2 < ii(n0) → lf(n0)[n2] = σ[n2], ∀ n2. n2 < # nextnli0(lf(n0), n0) → ((σ from ii(n0)) to # nextnli0(lf(n0), n0))[n2] = nextnli0(lf(n0), n0)[n2] ⊦ (lf(n0) + nextnli0(lf(n0), n0))[n1] = σ[n1], ii(n0) + # nextnli0(lf(n0), n0) = 0, # σ < ⌜ii(n0)⌝ + ⌜# nextnli0(lf(n0), n0)⌝

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: list-del inst: name: get-append-less seq: ⊦ n < # x → (x + y)[n] = x[n] subst: [n, y, x] → [n1, nextnli0(lf(n0), n0), lf(n0)] flags: rewrite

Additional information

Used terms: n1, nextnli0(lf(n0), n0), lf(n0)