Prove that A = B, regardless of what C is Prove which of the following statements imply A = B, regardless of what C is.
A - C = B - C
A ∩ C = B ∩ C
A ⋃ C = B ⋃ C
A△C = B△C
What might be wrong with the following approach?
if C = ∅:
A - C = B - C | → (A - ∅ = A; B - ∅ = B; so A = B);
A ∩ C = B ∩ C | → (A ∩ ∅ = ∅; B ∩ ∅ = ∅; so ∅ = ∅);
A ⋃ C = B ⋃ C | → (A ⋃ ∅ = A; B ⋃ ∅ = B, so A = B);
A △ C = B △ C | → (A - ∅) ⋃ (∅ - A) = A; (B - ∅) ⋃ (∅ - B) = B, so A=B;
else:
A - C = B - C | → (Solve for "A" →  A - C + C = B - C + C; so A = B);
A ∩ C = B ∩ C | → If ⊄, then let ∈∖, and ={}, so A ≠ B;
A ⋃ C = B ⋃ C | → ={1}, ={2}, ={1,2,3,4}; ∪={1,2,3,4}; ∪={1,2,3,4}; so A ≠ B;
A △ C = B △ C | → 1+1=1+1; 1+1+1=1+1+1; 1=1; so A=B;
 A: Another way of looking at the question, suppose that $A\neq B$.  For each statement, can we find a $C$ such that the equality holds?
For the first, by letting $C= A\cup B$ we have $A\setminus C = B\setminus C = \emptyset$ despite the fact that $A\neq B$.
For the second, by letting $C=\emptyset$ we have $A\cap C = B\cap C=\emptyset$ despite the fact that $A\neq B$.
For the third, by letting $C=A\cup B$ we have $A\cup C=B\cup C$ despite the fact that $A\neq B$.
Only the fourth, we cannot find such an example of $C$, and this can be proven as in the other answer.
A: the fourth is the only one who implies this equality regardless of what $C$ is, but you need to prove it, and not just bring examples:
because of symmetry, its enough to prove that $A \subseteq B$:
let $a \in A$, then lets split to cases:
case 1: $a \notin C$, therefor $a \in A \triangle C$. because $A \triangle C = B \triangle C$, $a \in B \triangle C$, and because $a \notin C$ its implies that $a \in B$, therefor, $A\subseteq B$.
case 2: $a \in C$, therefor $a \notin A \triangle C$. because $A \triangle C = B \triangle C$, $a \notin B \triangle C$, and because $a \in C$ its implies that $a \in B$, therefor, $A\subseteq B$.
as you shown, all the others are depends in what $C$ is and can't imply that $A=B$
