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Proof step 5 from proof restr-add-notin (previous step)

⊦ n ∈ gstore, ∀ n0. n0 ∈ (ns ++ n) \ dom gstore ↔ n0 ∈ (ns \ dom gstore) ++ n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ n0 ∈ (ns ++ n) \ dom gstore ↔ n0 = n ∨ n0 ∈ ns \ dom gstore, n ∈ gstore

Rule argument

Additional information

Used simplifier rules: ⊦ n ∈ ns ++ n0 ↔ n = n0 ∨ n ∈ ns; (s)