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Proof step 1 from proof in-06 (conclusion)

⊦ a ∈ y ∧ a ∈ x → a ∈ filter(rmdup(x), y)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: in-01-01: ⊦ a ∈ y → (a ∈ filter(x, y) ↔ a ∈ x);%rm-in-02: ⊦ a ∈ rmdup(x) ↔ a ∈ x; from specification list-dup