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Proof step 10 from proof restrict-union (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
@(m, m0, m1) ∈ as0 | n ∨ @(m, m0, m1) ∈ as1 | n ⊦ m = n ∧ @(n, m0, m1) ∈ as0 ∪ as1

Rule argument

Additional information

Used simplifier rules: restrict-in: ⊦ @(m, m0, m1) ∈ as | m ↔ @(m, m0, m1) ∈ as;%unc: ⊦ as ∪ as0 = as0 ∪ as; from specification set-union[address-store]%una: ⊦ (as ∪ as0) ∪ as1 = as ∪ as0 ∪ as1; from specification set-union[address-store]% ⊦ ¬ adr ∈ as → (adr ∈ as ∪ as0 ↔ adr ∈ as0); (s)