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Proof step 2 from proof unmodified-write-other (previous step)

⊦ unmodified(adr0, znd, iast[adr, ind]) ↔ unmodified(adr0, znd, iast), adr = adr0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: unmodified: ⊦ unmodified(adr, znd, iast) ↔ ¬ znd.dirty → adr ∈ iast ∧ iast[adr] = save(znd);% ⊦ adr ≠ adr0 → iast[adr0, ind][adr] = iast[adr]; (s)% ⊦ adr ≠ adr0 → (adr ∈ iast[adr0, ind] ↔ adr ∈ iast); (s)