Answer:
(a) and (b) are not true in general. Refer to the explanations below for counterexamples.
It can be shown that (c) is indeed true.
Step-by-step explanation:
This explanation will use a lot of empty sets [tex]\phi[/tex] just to keep the counterexamples simple.
Note that [tex]A \cap B[/tex] can well be smaller than [tex]A[/tex]. It should be alarming that the question is claiming [tex]A\![/tex] to be a subset of something that can be smaller than [tex]\! A[/tex]. Here's a counterexample that dramatize this observation:
Consider:
The intersection of an empty set with another set should still be an empty set:
[tex]A \cap B = \left\lbrace 1\right\rbrace \cap \left\lbrace\right\rbrace = \left\lbrace\right\rbrace[/tex].
The union of two empty sets should also be an empty set:
[tex]((A \cap B) \cup C) = \left\lbrace\right\rbrace \cup \left\lbrace\right\rbrace = \left\lbrace\right\rbrace[/tex].
Apparently, the one-element set [tex]A = \left\lbrace 1 \right\rbrace[/tex] isn't a subset of an empty set. [tex]A \not \subseteq ((A\cap B) \cup C)[/tex]. Contradiction.
Consider the same counterexample
Left-hand side:
[tex](A \cup B) \cap C = \left(\left\lbrace 1 \right\rbrace \cup \left\lbrace \right\rbrace\right) \cap \left\lbrace 2 \right\rbrace\right = \left\lbrace 1 \right\rbrace \cap \left\lbrace 2 \right\rbrace = \left\lbrace \right\rbrace[/tex].
Right-hand side:
[tex](A \cap B) \cup C = \left(\left\lbrace 1 \right\rbrace \cap \left\lbrace \right\rbrace\right) \cup \left\lbrace 2 \right\rbrace\right = \left\lbrace \right\rbrace \cup \left\lbrace 2 \right\rbrace = \left\lbrace 2 \right\rbrace[/tex].
Apparently, the empty set on the left-hand side [tex]\left\lbrace \right\rbrace[/tex] is not the same as the [tex]\left\lbrace 2 \right\rbrace[/tex] on the right-hand side. Contradiction.
Part one: show that left-hand side is a subset of the right-hand side.
Let [tex]x[/tex] be a member of the set on the left-hand side.
[tex]x \in (A \backslash B) \cap C[/tex].
[tex]\implies x\in A \backslash B[/tex] and [tex]x \in C[/tex] (the right arrow here reads "implies".)
[tex]\implies x \in A[/tex] and [tex]x \not\in B[/tex] and [tex]x \in C[/tex].
[tex]\implies (x \in A\cap C)[/tex] and [tex]x \not\in B \cap C[/tex].
[tex]\implies x \in (A \cap C) \backslash (B \cap C)[/tex].
Note that [tex]x \in (A \backslash B) \cap C[/tex] (set on the left-hand side) implies that [tex]x \in (A \cap C) \backslash (B \cap C)[/tex] (set on the right-hand side.)
Therefore:
[tex](A \backslash B) \cap C \subseteq (A \cap C) \backslash (B \cap C)[/tex].
Part two: show that the right-hand side is a subset of the left-hand side. This part is slightly more involved than the first part.
Let [tex]x[/tex] be a member of the set on the right-hand side.
[tex]x \in (A \cap C) \backslash (B \cap C)[/tex].
[tex]\implies x \in A \cap C[/tex] and [tex]x \not\in B \cap C[/tex].
Note that [tex]x \not\in B \cap C[/tex] is equivalent to:
However, [tex]x \in A \cap C[/tex] implies that [tex]x \in A[/tex] AND [tex]x \in C[/tex].
The fact that [tex]x \in C[/tex] means that the only possibility that [tex]x \not\in B \cap C[/tex] is [tex]x \not \in B[/tex].
To reiterate: if [tex]x \not \in C[/tex], then the assumption that [tex]x \in A \cap C[/tex] would not be true any more. Therefore, the only possibility is that [tex]x \not \in B[/tex].
Therefore, [tex]x \in (A \backslash B)\cap C[/tex].
In other words, [tex]x \in (A \cap C) \backslash (B \cap C) \implies x \in (A \backslash B)\cap C[/tex].
[tex](A \cap C) \backslash (B \cap C) \subseteq (A \backslash B)\cap C[/tex].
Combine these two parts to obtain: [tex](A \backslash B) \cap C = (A \cap C) \backslash (B \cap C)[/tex].
