1. Question on set operations

is this statement true?

(A - B) - C = (A - C) - (B - C)

2. Hello, brudman!

Is this statement true?

. . $(A - B) - C \:=\: (A - C) - (B - C)$ . . . . . yes
Definition: . $P - Q \:=\:P \cap Q'$

The left side is:

. . $\begin{array}{cc}
(A - B) - C & \text{Given} \\
(A \cap B') - C & \text{d{e}f.Subtr'n}\\
(A \cap B') \cap C' & \text{d{e}f.Subtr'n} \\
A \cap B' \cap C' & \text{Associative} \end{array}$

The right side is:

. . $\begin{array}{cc}
(A-C) - (B-C) & \text{Given} \\

(A \cap C') - (B \cap C') & \text{d{e}f.Subtr'n} \\

(A \cap C') \cap (B \cap C')' & \text{d{e}f.Subtr'n} \\

(A \cap C') \cap (B' \cup C) & \text{DeMorgan} \\

A \cap C' \cap (B' \cup C) & \text{Associative} \\

A \cap \bigg[(C' \cap B') \cup (C' \cap C)\bigg] & \text{Distributive} \\

A \cap \bigg[(C' \cap B') \cup \emptyset\bigg] & P \cap P \,=\,\emptyset
\end{array}$

. . . . . . $\begin{array}{ccccc}
A \cap (C' \cap B') &&&& P \cup \emptyset \,=\,P \\

A \cap (B' \cap C') &&&& \text{Commutative} \\

A \cap B' \cap C' &&&& \text{Associative}

\end{array}$