Prove that from Atriangle B = Ctriangle D it follows that A∪…

Prove that from $A\triangle B = C\triangle D$ it follows that $A\cup C = B\cup D$.

Step 1: The symmetric difference

\[ X\triangle Y = X\overline{Y} + \overline{X}Y = \text{"exactly one of } X, Y \text{ occurs"}. \]

For an outcome $\omega$, write $x=1$ if $\omega\in X$ and $x=0$ otherwise (similarly $y$). Then $\omega\in X\triangle Y$ exactly when $x+y$ is odd, i.e. when $x+y\equiv 1 \pmod 2$:

So $\triangle$ is "addition modulo $2$" of membership. It is commutative and associative, $X\triangle\varnothing=X$, and $X\triangle X=\varnothing$.

Step 2: The statement as written is false

Take $\Omega=\{1,2\}$ and

\[ A = C = \{1\}, \qquad B = D = \varnothing. \]

Then $A\triangle B = \{1\}$ and $C\triangle D = \{1\}$, so the assumption $A\triangle B=C\triangle D$ holds. But

\[ A\cup C = \{1\} \neq \varnothing = B\cup D. \]

One counterexample is enough: $A\triangle B=C\triangle D$ does not imply $A\cup C=B\cup D$. In words: $A$ and $C$ may both occur while $B$ and $D$ both fail — then "exactly one of $A,B$" and "exactly one of $C,D$" both occur, so the assumption is satisfied, yet $A\cup C$ occurs and $B\cup D$ does not.

Step 3: What the assumption really says

Take any outcome $\omega$ and let $a,b,c,d\in\{0,1\}$ record whether $\omega$ belongs to $A,B,C,D$. By Step 1,

\[ A\triangle B = C\triangle D \iff a+b \equiv c+d \pmod 2 \text{ for every } \omega \iff a+b+c+d \equiv 0 \pmod 2 \text{ for every } \omega. \]

(Adding $c+d$ to both sides and using $2c+2d\equiv 0$.) So:

\[ \boxed{A\triangle B = C\triangle D \iff \text{every outcome belongs to an even number (0, 2 or 4) of the events } A,B,C,D.} \]

This condition treats all four events symmetrically, so it can be regrouped in any way.

Step 4: The correct conclusion: $A\triangle C = B\triangle D$

Since $a+b+c+d\equiv 0$ for every outcome, also $a+c\equiv b+d \pmod 2$, i.e.

\[ \boxed{A\triangle B = C\triangle D \ \Longrightarrow\ A\triangle C = B\triangle D} \quad(\text{and also } A\triangle D = B\triangle C). \]

The same proof with the rules of $\triangle$

\[ A\triangle C = A\triangle(B\triangle B)\triangle C = (A\triangle B)\triangle(B\triangle C) = (C\triangle D)\triangle(B\triangle C) = (C\triangle C)\triangle D\triangle B = D\triangle B. \]

We inserted $B\triangle B=\varnothing$, regrouped (associativity), replaced $A\triangle B$ by $C\triangle D$, rearranged (commutativity) and used $C\triangle C=\varnothing$.

Step 5: Check on the counterexample

With $A=C=\{1\}$, $B=D=\varnothing$: $A\triangle C=\varnothing$ and $B\triangle D=\varnothing$ — equal, as the correct statement predicts. Outcome $1$ lies in exactly two of the events ($A$ and $C$), outcome $2$ in none: both even numbers. ✓

Step 6: A second example with a die

$\Omega=\{1,\dots,6\}$, $A=\{1,2,3\}$, $B=\{2,4\}$, $C=\{1,4,5\}$. Then $A\triangle B=\{1,3,4\}$. Choose $D$ so that $C\triangle D=\{1,3,4\}$: $D=C\triangle\{1,3,4\}=\{3,5\}$. Now

\[ A\triangle C=\{2,3,4,5\}, \qquad B\triangle D=\{2,3,4,5\}. \ \checkmark \qquad\text{but}\qquad A\cup C=\{1,2,3,4,5\}\ne B\cup D=\{2,3,4,5\}. \]