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Proof step 3 from proof Diff-union-comm (previous step)

⊦ ∀ key. (key ∈ (ri \ ks0) \ ks1 ↔ key ∈ ri \ (ks0 ∪ ks1)) ∧ (key ∈ (ri \ ks0) \ ks1 → ((ri \ ks0) \ ks1)[key] = (ri \ (ks0 ∪ ks1))[key])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information