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Proof step 6 from proof restrict-union-inter-empty-left (previous step)

⊦ adr ∈ as0 | n ∧ adr ∈ (as1 \ (as1 | n)) ∪ as2 ↔ adr ∈ (as0 | n) ∩ as2

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: restrict-leb seq: ⊦ adr ∈ as | n ↔ adr.lnum = n ∧ adr ∈ as subst: [n, as, adr] → [n, as0, adr] flags: <all paths>

Additional information

Used terms: n, as0, adr