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Proof step 4 from proof notin-add-diff-dom-nodes (previous step)

⊦ adr ∈ st, ∀ adr0. adr0 ∈ (as ++ adr) \ dom nodes(st) ↔ adr0 ∈ (as \ dom nodes(st)) ++ adr

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ adr0 ∈ (as ++ adr) \ dom nodes(st) ↔ adr0 = adr ∨ adr0 ∈ as \ dom nodes(st), adr ∈ st

Rule argument

Additional information

Used simplifier rules: ⊦ adr ∈ as ++ adr0 ↔ adr = adr0 ∨ adr ∈ as; (s)