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Proof step 2 from proof two-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

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