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Proof step 2 from proof in-06-01 (previous step)

⊦ dups(x + a ' + b ') ↔ a = b ∨ b ∈ x, dups(x + a ')

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: f: ⊦ dups(x + a ') ↔ a ∈ x ∨ dups(x);%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)