Prove by contradiction $a \in C$, if $a \in A \land a \not\in B \setminus C$

Question

This is the exercise I have:

Suppose that A ⊆ B, a ∈ A and a $\not\in$ B\C.

Prove by the method of contradiction that a ∈ C

Proving by contradiction means that if I find a contradiction trying to prove the opposite of what I want to prove, I prove what I want to prove.

So I have to suppose that $a \not\in C$.

$A \subset B$ is equivalent to say $\forall a (a \in A \rightarrow a \in B)$.

Then I know that $a \in A \land (a \not\in B \lor a \in C)$, where $(a \not\in B \lor a \in C)$ means $a \not\in A \setminus C$. But I know that $a \not\in C$, therefore I have $(a \not\in B \lor False)$, which is equivalent to $(a \not\in B)$, and in the end I have that $a \in A \land a \not\in B$, which is a contradiction of 1 of the premises, because an element of $A$ has to be also an element of $B$, from the first premise: $A \subset B$.

(I am sorry if I continue to ask for every exercise, but I would like to know if my proofs are correct, or at least to know if I am on the right track)

Answer 1

Suppose that $a \notin C$. Then since $a \in A \subseteq B, a\in B \Rightarrow a \in B$ and $a \notin C \Rightarrow a \in B$ \ $C$. Contradiction!

Answer 2

The statement $A \subseteq B$ means $\forall a(a \in A \rightarrow a \in B)$. The first statement that you wrote does not make sense. This implies that ANY $a$ is in both A and B. Not only does that mean that they are the same set, but that this set is the universal set.

Further note that $x \in B \setminus C$ means $x \in B \land x \notin C$ and so its negation: $x \notin B \setminus C$ means that either $x \notin B$ or $x \in C$.

Let $a \in A$. Suppose $a \notin C$. Then $a \notin B$ However, by property of subset $a \in A \rightarrow a \in B$. This is a contradiction.

This should work...