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Proof step 3 from proof in-03 (previous step)

⊦ a ∈ rmdup((x + y) + z) ↔ a ∈ rmdup(x + z), a ∈ y

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

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