Show that the statements $C \subseteq A \cup B$ and $C \setminus A \subseteq B$ are equivalent. "Show that the statements $C \subseteq (A \cup B)$ and $C \setminus (A \subseteq B)$ are equivalent by writing each in logical symbols and then showing that the resulting formulas are equivalent."
I can't seem to figure this one out at all. Any help would be appreciated.
 A: I agree with Fred in the comments, the second expression shouldn’t be $C \setminus (A \subseteq B)$ but $(C \setminus A) \subseteq B.$ Here’s why.
Applying the definitions of each set operation, we get the following logic expressions.
$$C \subseteq (A \cup B) \iff (\forall x):[x \in C \to (x \in A \vee x \in B)].$$
$$C \setminus (A \subseteq B) \iff (\forall x):[x \in C \wedge \neg (x \in A \to x \in B)].$$
Let’s see why the second is not equivalent to the first.

*

*$C \subseteq (A \cup B) \iff (\forall x):[x \in C \to (x \in A \vee x \in B)] \iff$
$ \iff (\forall x):[x \notin C \vee x \in A \vee x \in B].$


*$(\forall x):[x \in C \wedge \neg (x \in A \to x \in B)]$ $\iff (\forall x):[x \in C \wedge \neg (x \notin A \vee x \in B)]$
$ \iff (\forall x):[x \in C \wedge x \in A \wedge x \notin B].$
Suppose that $x \in B.$ Then, for this element, $1.$ would be true and $2.$ would be false, so they are not equivalent.
Although, writing the second expression as $(C \setminus A) \subseteq B$ we get an equivalence.
$$(C \setminus A) \subseteq B \iff (\forall x):[x \in C \setminus A \to x \in B] \iff (\forall x):[x \notin C \setminus A \vee x \in B] \iff (\forall x):[(x \notin C \vee x \in A) \vee x \in B] \iff (\forall x):[(x \notin C) \vee (x \in A \vee x \in B)].$$
Using the Material Implication, we see that the last expression is equivalent to
$$(\forall x):[x \in C \to (x \in A \vee x \in B)].$$
Which is equivalent to the logic form of the first expression, therefore $C \subseteq (A \cup B)$ and $(C \setminus A) \subseteq B$ are equivalent.
