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Proof step 11 from proof diff-subseteq-subseteq (previous step)

as1 ⊆ as2, ∀ adr. adr ∈ as0 → adr ∈ as2 ⊦ ∀ adr. adr ∈ as0 \ as1 → adr ∈ as2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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]% ⊦ adr ∈ as \ as0 ↔ adr ∈ as ∧ ¬ adr ∈ as0; (ws)