Distributive Laws/Set Theory/Examples/A cap B cap (C cup D) subset of (A cap D) cup (B cap C)/Corollary
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Corollary to $A \cap B \cap \paren {C \cup D} \subseteq \paren {A \cap D} \cup \paren {B \cap C}$
Let:
- $P = A \cap B \cap \paren {C \cup D}$
- $Q = \paren {A \cap D} \cup \paren {B \cap C}$
Then:
- $P = Q$
- both $B \cap C \subseteq A$ and $A \cap D \subseteq B$
Proof
Sufficient Condition
Let $P = Q$.
\(\ds B \cap C\) | \(\subseteq\) | \(\ds \paren {A \cap D} \cup \paren {B \cap C}\) | Set is Subset of Union | |||||||||||
\(\ds \) | \(=\) | \(\ds Q\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds P\) | by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds A \cap B \cap \paren {C \cup D}\) | by hypothesis | |||||||||||
\(\ds \) | \(\subseteq\) | \(\ds A\) | Intersection is Subset |
and:
\(\ds A \cap D\) | \(\subseteq\) | \(\ds \paren {A \cap D} \cup \paren {B \cap C}\) | Set is Subset of Union | |||||||||||
\(\ds \) | \(=\) | \(\ds Q\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds P\) | by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds A \cap B \cap \paren {C \cup D}\) | by hypothesis | |||||||||||
\(\ds \) | \(\subseteq\) | \(\ds B\) | Intersection is Subset |
Thus we have both:
- $B \cap C \subseteq A$
and
- $A \cap D \subseteq B$
$\Box$
Necessary Condition
Let $B \cap C \subseteq A$ and $A \cap D \subseteq B$.
We have:
\(\ds B \cap C\) | \(\subseteq\) | \(\ds A\) | by hypothesis | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds B \cap C\) | \(=\) | \(\ds A \cap B \cap C\) | Intersection with Subset is Subset |
and:
\(\ds A \cap D\) | \(\subseteq\) | \(\ds B\) | by hypothesis | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds A \cap D\) | \(=\) | \(\ds A \cap B \cap D\) | Intersection with Subset is Subset |
and so:
\(\ds \paren {A \cap D} \cup \paren {B \cap C}\) | \(=\) | \(\ds \paren {A \cap B \cap C} \cup \paren {A \cap B \cap D}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds A \cap B \cap \paren {C \cup D}\) | Intersection Distributes over Union |
and so $P = Q$ as required.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $1$: Sets and Logic: Exercise $11$