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Proof step 8 from proof restrict-2 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
adr ∈ as1, adr ∈ as0, adr ∈ iast ⊦ # as0 = 0, # as1 = 0, iast[adr] = (iast | (as0 ∩ as1))[adr], as0 = ∅, as1 = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ adr ∈ iast | as ↔ adr ∈ iast ∧ adr ∈ as; (ws)% ⊦ adr ∈ iast ∧ adr ∈ as → (iast | as)[adr] = iast[adr]; (s)