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Proof step 3 from proof restrict-2 (previous step)

⊦ ∀ adr. (adr ∈ (iast | as0) | as1 ↔ adr ∈ iast | (as0 ∩ as1)) ∧ ((iast | as0) | as1)[adr] = (iast | (as0 ∩ as1))[adr]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (adr ∈ (iast | as0) | as1 ↔ adr ∈ iast | (as0 ∩ as1)) ∧ ((iast | as0) | as1)[adr] = (iast | (as0 ∩ as1))[adr]

Rule argument

Additional information