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Proof step 99 from proof infinite-append-prefixnextnli (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 100:
⊦ x ⊑ y ↔ (∃ z. x + z = y)
Proof step 101:
∃ z. x + z = nextnli(x, n) ⊦ ∃ y. nextnli(x, n) = x + y, ∃ σ, ii. ii(0) = 0 ∧ (∀ n. ii(n) ≤ ii(n + 1) ∧ ¬ # σ < ⌜ii(n + 1)⌝ ∧ σ to ii(n + 1) = ⌜nextnli((σ to ii(n)).l, n)⌝ ) ∧ (∀ m, n. m ≤ n → ii(m) ≤ ii(n)) ∧ (finp(σ) ⊃ ∃ n. # σ = ⌜ii(n)⌝ ∧ ⌜nextnli(σ.l, n)⌝ = σ ∧ (∀ m. n ≤ m → ii(m) = ii(n));∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1))

Rule argument

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

Additional information

Used terms: nextnli(x, n), x