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Proof step 4 from proof firstn-restn-in (previous step)

n0 < # x, ∀ x, a. n0 ≤ # x → (a ∈ x ↔ a ∈ firstn(n0, x) ∨ a ∈ restn(n0, x)) ⊦ a ∈ x ↔ a = x.first ∨ a ∈ firstn(n0, x.rest) ∨ a ∈ restn(n0, x.rest), x = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
x = a0 + y, n0 < # x, ∀ x, a. n0 ≤ # x → (a ∈ x ↔ a ∈ firstn(n0, x) ∨ a ∈ restn(n0, x)) ⊦ a ∈ x ↔ a = a0 ∨ a ∈ firstn(n0, y) ∨ a ∈ restn(n0, y), x = []

Rule argument

spec: list-data inst: name: elim-list-c seq: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y subst: [x, a, y] → [x, a0, y] flags:

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: x, a0, y