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Proof step 3 from proof restrict-del (previous step)

⊦ ∀ adr. adr ∈ (as -- @(m, m0, m1)) | m ↔ adr ∈ (as | m) -- @(m, m0, m1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ adr ∈ (as -- @(m, m0, m1)) | m ↔ adr ≠ @(m, m0, m1) ∧ adr ∈ as | m

Rule argument

Additional information

Used simplifier rules: ⊦ adr ∈ as -- adr0 ↔ adr ≠ adr0 ∧ adr ∈ as; (s)