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Theorem

Let $ A $ be a set in S. Then
A ∪ S = S.



Proof

Let x ∈ A ∪ S. Then x ∈ A or x ∈ S.
If x ∈ A, then x ∈ S since A ⊆ S.
If x ∈ S, then, well, x ∈ S.
So we have that if x ∈ A ∪ S, then x ∈ S ⇒ A ∪ SS.

Next, we want to show that S ⊆ A ∪ S.
We know this is true because the set resulting from the union of two sets contains both of the sets forming the union (proof).

Since A ∪ SS and S ⊂ A ∪ S, we have that A ∪ S = S.
$ \blacksquare $



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