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Proof step 4 from proof firstN-dups-04 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

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

Additional information

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