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Proof step 3 from proof rev-dups (previous step)

dups(rev(x0)) ↔ dups(x0) ⊦ dups(rev(x0) + a ') ↔ a ∈ x0 ∨ dups(x0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ dups(x + a ') ↔ a ∈ x ∨ dups(x)

Rule argument

spec: list-dup inst: name: f seq: ⊦ dups(x + a ') ↔ a ∈ x ∨ dups(x) subst: [a, x] → [a, rev(x0)] flags: rewrite

Additional information

Used simplifier rules: rev-in: ⊦ a ∈ rev(x) ↔ a ∈ x; Used terms: a, rev(x0)