# Set Difference with Union is Set Difference

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## Theorem

The set difference between two sets is the same as the set difference between their union and the second of the two sets:

Let $S, T$ be sets.

Then:

- $\left({S \cup T}\right) \setminus T = S \setminus T$

## Proof

Consider $S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universe.

\(\displaystyle \left({S \cup T}\right) \setminus T\) | \(=\) | \(\displaystyle \left({S \cup T}\right) \cap \overline T\) | Set Difference as Intersection with Complement | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({S \cap \overline T}\right) \cup \left({T \cap \overline T}\right)\) | Intersection Distributes over Union | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({S \setminus T}\right) \cup \left({T \setminus T}\right)\) | Set Difference as Intersection with Complement | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({S \setminus T}\right) \cup \varnothing\) | Set Difference with Self is Empty Set | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle S \setminus T\) | Union with Empty Set |

$\blacksquare$