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## Main Question or Discussion Point

Below is a attempt at a proof of a distributive property of union/intersections of sets. A critique would be very much appreciated. Thanks in advance!

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For sets A, B, and C, the formula,

##A \cup (B \cap C) = (A \cup B) \cap (A \cup C)##

is true

Two sets are equal if they contain the same elements. Let x denote an element of set ##A \cup (B \cap C)## and y an element of ##(A \cup B) \cap (A \cup C)##. Where X and Y are sets, ##X \subset Y## and ##e \in Y##, define the function ##f_X(e)## as

##f_X(e) =

\begin{cases}

1, & e \in X \\

0, & e \notin X

\end{cases}##

We want to show that ##\forall e \in A \cup B \cup C, f_{A \cup (B \cap C)}(e) = f_{(A \cup B) \cap (A \cup C)}(e)##, since this would imply ##e \in A \cup (B \cap C) \iff e \in (A \cup B) \cap (A \cup C)##. First, two basic and easily verifiable properties of ##f_X(e)## are:

##\begin{alignat}{0}

f_{X \cap Y}(e) = f_X(e)f_Y(e) & \text{(1) - since e must be in X and Y} \\

f_{X \cup Y}(e) = f_X(e) + f_Y(e) - f_X(e)f_Y(e) & \text{(2) - since e may be in X or Y}

\end{alignat}##

And now:

##f_{A \cup (B \cap C)}(e)##

##

\begin{alignat}{0}

=&f_A(e) + f_{B \cap C}(e) - f_A(e)f_{B \cap C}(e) & \text{(1)}\\

=&f_A(e) + f_B(e) f_C(e) - f_A(e)f_B(e)f_C(e) & \text{(0)}\\

=&f_{(A \cup B) \cap (A \cup C)}& \text{(1)}

\end{alignat}

##

And therefore ##e \in A \cup (B \cap C) \iff e \in (A \cup B) \cap (A \cup C)## and ##A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \text{ } \Box##

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P.S.: This is not a homework assignment; however, it is very much

-----------------

For sets A, B, and C, the formula,

##A \cup (B \cap C) = (A \cup B) \cap (A \cup C)##

is true

Two sets are equal if they contain the same elements. Let x denote an element of set ##A \cup (B \cap C)## and y an element of ##(A \cup B) \cap (A \cup C)##. Where X and Y are sets, ##X \subset Y## and ##e \in Y##, define the function ##f_X(e)## as

##f_X(e) =

\begin{cases}

1, & e \in X \\

0, & e \notin X

\end{cases}##

We want to show that ##\forall e \in A \cup B \cup C, f_{A \cup (B \cap C)}(e) = f_{(A \cup B) \cap (A \cup C)}(e)##, since this would imply ##e \in A \cup (B \cap C) \iff e \in (A \cup B) \cap (A \cup C)##. First, two basic and easily verifiable properties of ##f_X(e)## are:

##\begin{alignat}{0}

f_{X \cap Y}(e) = f_X(e)f_Y(e) & \text{(1) - since e must be in X and Y} \\

f_{X \cup Y}(e) = f_X(e) + f_Y(e) - f_X(e)f_Y(e) & \text{(2) - since e may be in X or Y}

\end{alignat}##

And now:

##f_{A \cup (B \cap C)}(e)##

##

\begin{alignat}{0}

=&f_A(e) + f_{B \cap C}(e) - f_A(e)f_{B \cap C}(e) & \text{(1)}\\

=&f_A(e) + f_B(e) f_C(e) - f_A(e)f_B(e)f_C(e) & \text{(0)}\\

=&f_{(A \cup B) \cap (A \cup C)}& \text{(1)}

\end{alignat}

##

And therefore ##e \in A \cup (B \cap C) \iff e \in (A \cup B) \cap (A \cup C)## and ##A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \text{ } \Box##

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P.S.: This is not a homework assignment; however, it is very much

*like*a homework assignment. Would this type of problem go better in the homework section?