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Proof step 5 from proof restrict-2 (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
adr ∈ as0, adr ∈ iast, adr ∈ as1 ⊦ # as0 = 0, # as1 = 0, ((iast | as0) | as1)[adr] = (iast | (as0 ∩ as1))[adr], as0 = ∅, as1 = ∅

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ as ≠ ∅ → # as ≠ 0; (forward) from specification set-basic[node-store]%ne-01: ⊦ adr ∈ as → as ≠ ∅; (forward) from specification set-basic[node-store]%int-c: ⊦ as ∩ as0 = as0 ∩ as; from specification set-union[node-store]%int-a: ⊦ (as ∩ as0) ∩ as1 = as ∩ as0 ∩ as1; from specification set-union[node-store]% ⊦ adr ∈ as ∩ as0 ↔ adr ∈ as ∧ adr ∈ as0; (ws)% ⊦ adr ∈ iast | as ↔ adr ∈ iast ∧ adr ∈ as; (ws)