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Proof step 15 from proof diff-eq-diff-eq-union (previous step)

as0 ⊆ as2, as1 ⊆ as2, as0 ⊆ as1, adr ∈ as2 ∧ ¬ adr ∈ as1 ∨ adr ∈ as1 ∧ ¬ adr ∈ as0 ⊦ adr ∈ as2 ∧ ¬ adr ∈ as0

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 4

Additional information

Used simplifier rules: sub-in: ⊦ adr ∈ as ∧ as ⊆ as0 → adr ∈ as0; (forward) from specification set-basic[node-store]%ne-01: ⊦ adr ∈ as → as ≠ ∅; (forward) from specification set-basic[node-store]