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Proof step 3 from proof at-put (previous step)

⊦ a = b, ([])[b, d][a] = ([])[a]

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: put-empty: ⊦ ([])[a, d] = (a × d)';%lookup-one: ⊦ a ≠ b → (b × d)'[a] = ([])[a];