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Proof step 20 from proof referenced-adrs-in (previous step)

adr ∈ referenced-adrs(ax0, ri), ∀ adr, ri. adr ∈ referenced-adrs(ax0, ri) ↔ adr ∈ ax0 ∧ adr ∈ ri ⊦ adr0 ∈ ri, adr ∈ ax0, adr ∈ ax0 ∧ adr ∈ ri

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information