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

⊦ a ∈ b ' ∪ y ↔ a ∈ y, a = b

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: union: ⊦ x ∪ y = rmdup(x + y);%assoc: ⊦ (x ∪ y) ∪ z = x ∪ y ∪ z;%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ rmdup(x + y) ↔ a ∈ x ∨ a ∈ y; (ws)