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Proof step 196 from proof infinite-append-prefixnextli (previous step)

x ⊑ nextli(x) ⊦ ∃ y. nextli(x) = x + y, ∃ σ, ii. ii(0) = 0 ∧ (∀ n. ¬ # σ < ⌜ii(n + 1)⌝ ∧ σ to ii(n + 1) = ⌜nextli((σ to ii(n)).l)⌝ ∧ (# σ = ⌜ii(n)⌝ ⊃ ii(n) = ii(n + 1);ii(n) < ii(n + 1))) ∧ (∀ m, n. m ≤ n → ii(m) ≤ ii(n)) ∧ (finp(σ) ⊃ ∃ n. # σ = ⌜ii(n)⌝ ∧ ⌜nextli(σ.l)⌝ = σ ∧ (∀ m. (n ≤ m ⊃ ii(m) = ii(n);ii(m) < ii(m + 1))) ;∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1) )

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: list-last inst: name: prefix seq: ⊦ x ⊑ y ↔ (∃ z. x + z = y) subst: [y, x] → [nextli(x), x] flags: rewrite

Additional information

Used simplifier rules: app-01: ⊦ x + y = x + z ↔ y = z; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: nextli(x), x