Proof of $A\times(B-C)=(A\times B)-(A\times C)$ For any three sets $A,B,C$ I have to prove $A\times(B-C)=(A\times B)-(A\times C)$.
My book shows the following proof-

Let,arbitrary $(a,b)\in A\times(B-C)$
$\implies a\in A$ and $\color{blue}{b\in(B-C)}$
$\implies a\in A$ and $(b\in B$ and $b\not\in C)$
$\implies (a\in A$ and $\color{red}{b\in B})$ and $(a\in A$ and $b\not\in C)$
$\implies(a,b)\in(A\times B)$ and $(a,b)\notin(A\times C)$
$\implies(a,b)\in(A\times B)-(A\times C)$.

Now my question is in the part marked red we assume $b\in(B-C)$ which means $b\in B$ and $b\notin C$.But in the red part we say $b\in B$ which means we include the whole set $B$ without excluding the intersection with $C$ part.
Now does these two things not contradict each other?I have a doubt whether this proof is correct.
Edit-No reason why my post as closed as duplicate.I didn't want the proof.I specifically had difficulty with a particular logic.So no point redirecting my question to another proof without explaining my difficulty specifically.I already have that proof in my book.
 A: For the marked part you can simply notice that $p\land (q\land r) \Leftrightarrow (p\land q) \land (p\land r)$ is a tautology. 
Now substitute $p\equiv (a\in A)$, $q\equiv(b\in B)$ and $r\equiv(b\notin C)$.
A: The proof is fine.  By saying that $b \in B$ we are not making any claims whether or not $b$ is in $C$ or not.  But given that $b \in B - C$, it is definitely true that $b \in B$.
Think about it this way. Since $b \in B - C$, we know two things: $b \in B$ and $b \not \in C$.  What logically follows? It logically follows that $b \in B$. That's what the $\Rightarrow$ indicates.
Put differently: the claim $b \in B$ is equivalent to the claim $b \in B$ and either $b \in C$ or $b \not \in C$.  And that claim logically follows from $b \in B$ and $b \not \in C$.
Put differently yet: from the claim that 'It rains and today is Friday', we can infer the claim 'it rains'.  The fact that the latter claim says nothing about today being Friday or not does not take anything away from that logical consequence.
