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Proof step 2 from proof in-03 (previous step)

⊦ a ∈ x ∪ y ∪ z ↔ a ∈ x ∪ z, a ∈ y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ a ∈ rmdup((x + y) + z) ↔ a ∈ rmdup(x + z), a ∈ y

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% ⊦ rmdup(rmdup(x) + y) = rmdup(x + y); (s)