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Proof step 8 from proof dups-01 (previous step)

⊦ dups(x), ∃ a. a ∈ x ∧ (a ∈ s ∨ a ∈ s0), (¬ a ∈ s ∨ ¬ a ∈ x) ∧ ¬ dups(x + set2list(s0))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ dups(x + y) ↔ dups(x) ∨ dups(y) ∨ (∃ a. a ∈ x ∧ a ∈ y)
Proof step 10:
⊦ dups(x), ∃ a. a ∈ x ∧ (a ∈ s ∨ a ∈ s0), (¬ a ∈ s ∨ ¬ a ∈ x) ∧ ¬(dups(x) ∨ dups(set2list(s0)) ∨ (∃ a. a ∈ x ∧ a ∈ set2list(s0)))

Rule argument

seq: ⊦ dups(x + y) ↔ dups(x) ∨ dups(y) ∨ (∃ a. a ∈ x ∧ a ∈ y) subst: [y, x] → [set2list(s0), x] rotate: F spec: list-dup inst: name: dups-appex

Additional information

Used terms: set2list(s0), x