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Proof step 1 from proof l-03 (conclusion)

⊦ ¬ b ∈ x → (dups(x + b ') ↔ dups(x))

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: f: ⊦ dups(x + a ') ↔ a ∈ x ∨ dups(x);%l-02: ⊦ dups(x) → dups(x + b ');%a: ⊦ (x + y) + z = x + y + z; from specification list