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Proof step 5 from proof inter-subseteq-simp (previous step)

as0 ∩ as ⊆ as1 ⊦ as1 ⊆ as, ∀ adr. adr ∈ as0 ∩ as → adr ∈ as1 ∩ as

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
adr ∈ as, adr ∈ as0, as0 ∩ as ⊆ as1 ⊦ # as = 0, # as0 = 0, as = ∅, as0 = ∅, adr ∈ as1, as1 ⊆ as

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ as ≠ ∅ → # as ≠ 0; (forward) from specification set-basic[node-store]%ne-01: ⊦ adr ∈ as → as ≠ ∅; (forward) from specification set-basic[node-store]%int-c: ⊦ as ∩ as0 = as0 ∩ as; from specification set-union[node-store]%int-a: ⊦ (as ∩ as0) ∩ as1 = as ∩ as0 ∩ as1; from specification set-union[node-store]% ⊦ adr ∈ as ∩ as0 ↔ adr ∈ as ∧ adr ∈ as0; (ws)