Equal Set Differences iff Equal Intersections

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Theorem

$R \setminus S = R \setminus T \iff R \cap S = R \cap T$


Proof 1

\(\displaystyle R \setminus S\) \(=\) \(\displaystyle R \setminus T\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \set {x \in R: x \notin S}\) \(=\) \(\displaystyle \set {x \in R: x \notin T}\) Definition of Set Difference
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \forall x \in R: \quad x \notin S\) \(\iff\) \(\displaystyle x \notin T\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \forall x \in R: \quad x \in S\) \(\iff\) \(\displaystyle x \in T\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \set {\paren {x \in R} \land \paren {x \in S} }\) \(=\) \(\displaystyle \set {\paren {x \in R} \land \paren {x \in T} }\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle R \cap S\) \(=\) \(\displaystyle R \cap T\) Definition of Set Intersection

$\blacksquare$


Proof 2

From Set Difference and Intersection form Partition:

$\paren {R \setminus S} \cup \paren {R \cap S} = R = \paren {R \setminus T} \cup \paren {R \cap T}$
$\paren {R \cap S} \cap \paren {R \setminus S} = \O = \paren {R \cap T} \cap \paren {R \setminus T}$

whatever $R, S, T$ might be.


Let $R \setminus S = R \setminus T$.

Then:

\(\displaystyle \paren {\paren {R \setminus S} \cup \paren {R \cap S} } \setminus \paren {R \setminus S}\) \(=\) \(\displaystyle \paren {\paren {R \setminus T} \cup \paren {R \cap T} } \setminus \paren {R \setminus T}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {R \cap S} \setminus \paren {R \setminus S}\) \(=\) \(\displaystyle \paren {R \cap T} \setminus \paren {R \setminus T}\) Set Difference with Union is Set Difference


Now, we have from Set Difference with Disjoint Set:

$S \cap T = \O \iff S \setminus T = S$

and so:

$\paren {R \cap S} \setminus \paren {R \setminus S} = R \cap S$

and:

$\paren {R \cap T} \setminus \paren {R \setminus T} = R \cap T$

So:

$R \cap S = R \cap T$


We can use exactly the same reasoning if we assume $R \cap S = R \cap T$:

\(\displaystyle \paren {\paren {R \setminus S} \cup \paren {R \cap S} } \setminus \paren {R \cap S}\) \(=\) \(\displaystyle \paren {\paren {R \setminus T} \cup \paren {R \cap T} } \setminus \paren {R \cap T}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {R \setminus S} \setminus \paren {R \cap S}\) \(=\) \(\displaystyle \paren {R \setminus T} \setminus \paren {R \cap T}\) Set Difference with Union is Set Difference

and then because of Set Difference with Disjoint Set as above:

$\paren {R \setminus S} \setminus \paren {R \cap S} = R \setminus S$

and:

$\paren {R \setminus T} \setminus \paren {R \cap T} = R \setminus T$


So:

$R \setminus S = R \setminus T$

$\blacksquare$


Sources