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Proof step 1 from proof subseteq-union-diff-one (conclusion)

n < # asar, n0 < # asar ⊦ disj(asar, 0, # asar) → (asar[n] ⊆ (union(asar, 0, # asar) \ asar[n0]) ∪ as ↔ n ≠ n0 ∨ asar[n0] ⊆ as)

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information