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Proof step 8 from proof inter-union-diff-subseteq (previous step)

adr ∈ as3, adr ∈ as0, as1 ∩ as3 ⊆ as, ∀ adr. adr ∈ (as0 ∪ (as1 \ as2)) ∩ as3 → adr ∈ as ⊦ # as3 = 0, # as0 = 0, as0 = ∅, as3 = ∅, adr ∈ as

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: adr

Additional information

Used simplifier rules: ⊦ adr ∈ as ∩ as0 ↔ adr ∈ as ∧ adr ∈ as0; (ws)%int-a: ⊦ (as ∩ as0) ∩ as1 = as ∩ as0 ∩ as1; from specification set-union[node-store]%int-c: ⊦ as ∩ as0 = as0 ∩ as; from specification set-union[node-store]% ⊦ adr ∈ as ∪ as0 ↔ ¬(¬ adr ∈ as ∧ ¬ adr ∈ as0); (ss)%una: ⊦ (as ∪ as0) ∪ as1 = as ∪ as0 ∪ as1; from specification set-union[node-store]%unc: ⊦ as ∪ as0 = as0 ∪ as; from specification set-union[node-store] Used terms: adr