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Proof step 5 from proof restrict-add-other (previous step)

⊦ m = m0, ∀ adr. adr ∈ (as ++ @(m0, m1, m2)) | m ↔ adr ∈ as | m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ adr ∈ (as ++ @(m0, m1, m2)) | m ↔ adr ∈ as | m, m = m0

Rule argument

Additional information