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Proof step 2 from proof toSet-diff-in (previous step)

⊦ key ∈ ri \ kx.toSet ↔ key ∈ ri, key ∈ kx

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ key ∈ kx.toSet ↔ key ∈ kx; (s)% ⊦ key ∈ ri \ ks ↔ key ∈ ri ∧ ¬ key ∈ ks; (ws)