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Proof step 3 from proof union-diff-same (
previous step
)
⊦ ∀ adr. adr ∈ as0 ∪ (as1 \ as0) ↔ adr ∈ as0 ∪ as1
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 4
:
⊦ adr ∈ as0 ∪ (as1 \ as0) ↔ adr ∈ as0 ∪ as1
Rule argument
Additional information