Prove that $(A×B)\setminus(C×C)= ((A\setminus C) ×B) ∪ (A×(B\setminus C)$ Let $A, B$ and $C$ be sets, with $C\subseteq A\cap B$. Prove that 

$$(A×B)\setminus(C×C)= ((A\setminus C) ×B) ∪ (A×(B\setminus C))$$
  MY try:
  $$∃(x,y):(x,y)∈(A\setminus C) ×B ∨ A×(B\setminus C)≡$$$$(x∈(A\setminus C) ∧ y∈B ) ∨ ( x∈A ∧ y∈(B\setminus C)≡$$$$(x∈A ∧ x∉C ∧ y∈B) ∨ ( 
x∈A ∧ y∈B ∧ y∉C)≡$$$$((x,y )∈A×B ∧ x∉C) ∨ ( (x,y )∈A×B ∧ y∉C)≡$$$$((x,y )∈A×B ∧(x,y )∈A×B) ∨ (x∉C ∧
y∉C)≡$$$$((x,y)∈A×B ) ∨ ((x,y)∉C×C )$$
  But this is not the right answer, so what should I do?

 A: The first line of your 'proof' begins with $\exists(x,y)$... this is not a good way to start an elementary equality of sets proof. Your proof should go something like this:
"Suppose that $(x,y)\in(A\times B)\setminus(C\times C)$. Then ... [insert proof here] ... and therefore $(x,y)\in((A\setminus C)\times B)\cup(A\times(B\setminus C))$. Now suppose that $(x,y)\in((A\setminus C)\times B)\cup(A\times(B\setminus C))$. Then ... [insert proof here] ... and therefore $(x,y)\in(A\times B)\setminus(C\times C)$."
This is the structure you're aiming for. And each line needs to follow clearly from the last simply by the definitions of each set. Also I've noticed a problem some small but rectifiable problems in the last two lines of what you've written: There's a redundancy on the second last line (can you spot what it is), and on both the last two lines you write $\vee$ where you should have written $\wedge$.
Other than that what you've written is good, it just needs some words explaining what you're actually doing to make it a proof. Hopefully that helps.
A: Starting from your third row:
$$[(x \in A \wedge y \in B) \wedge (x \notin C)] \vee [(x \in A \wedge y \in B) \wedge (y \notin C)]=$$$$=(x \in A \wedge y \in B) \wedge [(x \notin C)\vee (y \notin C)]$$ by the distributive law.
$(x \in A \wedge y \in B) \wedge [(x \notin C)\vee (y \notin C)] = (x \in A \wedge y \in B) \wedge [\neg((x \in C)\wedge (y \in C))]$
$$\Rightarrow (x,y) \in A\times B \setminus  (C\times C)$$
