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Proof step 2 from proof inter-notin-add-empty (previous step)

⊦ (ns ++ n) ∩ dom istore = ∅ ↔ ns ∩ dom istore = ∅, n ∈ istore

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ n ∈ dom istore ↔ n ∈ istore; (s)% ⊦ ns ∩ (ns0 ++ n) = ∅ ↔ ns ∩ ns0 = ∅ ∧ ¬ n ∈ ns; (ws)%int-c: ⊦ ns ∩ ns0 = ns0 ∩ ns; from specification set-union[nat-set]%int-a: ⊦ (ns ∩ ns0) ∩ ns1 = ns ∩ ns0 ∩ ns1; from specification set-union[nat-set]