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Proof step 11 from proof restrict-union (previous step)

@(m, m0, m1) ∈ as0 | n ∨ @(m, m0, m1) ∈ as1 | n ⊦ m = n ∧ @(n, m0, m1) ∈ as0 ∪ as1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ adr ∈ as ∪ as0 ↔ adr ∈ as ∨ adr ∈ as0
Proof step 13:
@(m, m0, m1) ∈ as0 | n ∨ @(m, m0, m1) ∈ as1 | n ⊦ m = n ∧ (@(n, m0, m1) ∈ as0 ∨ @(n, m0, m1) ∈ as1)

Rule argument

spec: set-union, inst: address-store name: Union seq: ⊦ adr ∈ as ∪ as0 ↔ adr ∈ as ∨ adr ∈ as0 subst: [as0, as, adr] → [as1, as0, @(n, m0, m1)] flags: <all paths>

Additional information

Used terms: as1, as0, @(n, m0, m1)