Distributive Laws/Set Theory/Examples/A cap B cap (C cup D) subset of (A cap D) cup (B cap C)/Corollary

From ProofWiki
Jump to navigation Jump to search

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:

$P = Q$

if and only if:

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

Retrieved from "https://proofwiki.org/w/index.php?title=Distributive_Laws/Set_Theory/Examples/A_cap_B_cap_(C_cup_D)_subset_of_(A_cap_D)_cup_(B_cap_C)/Corollary&oldid=669776"