I would like to know if my procedure was correct in proving the next property ($\oplus \equiv$ symmetric difference):

$$(A_1\cup A_2)\oplus (B_1\cup B_2)\subset (A_1\oplus B_1)\cup (A_2\oplus B_2)$$

$$Let\ x\in(A_1\cup A_2)\oplus (B_1\cup B_2)$$

$$i.e.\ x\in(A_1\cup A_2)\setminus (B_1\cup B_2)\ \lor\ x\in(B_1\cup B_2)\setminus(A_1\cup A_2)$$

$$i.e.\ [x\in(A_1\cup A_2)\ \land\ x\notin(B_1\cup B_2)]\ \lor\ [x\in(B_1\cup B_2)\ \land\ x\notin(A_1\cup A_2)]$$

$$i.e.\ [(x\in A_1\lor x\in A_2)\land(x\notin B_1\land x\notin B_2)]\ \lor\ [(x\in B_1\lor x\in B_2)\land(x\notin A_1\land x\notin A_2)]$$

$$i.e.\ (x\in A_1\land x\notin B_1\land x\notin B_2)\lor(x\in A_2\land x\notin B_1\land x\notin B_2)\ \lor\ (x\in B_1\land x\notin A_1\land x\notin A_2)\lor (x\in B_2\land x\notin A_1\land x\notin A_2)$$

so, particularly, we have

$$\Rightarrow (x\in A_1\land x\notin B_1)\lor(x\in A_2\land x\notin B_2)\lor(x\in B_1\land x\notin A_1)\lor(x\in B_2\land x\notin A_2)$$

and, since $\lor$ is asociative, we have

$$\Rightarrow ((x\in A_1\land x\notin B_1)\lor(x\in B_1\land x\notin A_1))\lor((x\in A_2\land x\notin B_2)\lor(x\in B_2\land x\notin A_2))$$

$$i.e.\ (x\in A_1\setminus B_1\ \lor\ x\in B_1\setminus A_1)\lor(x\in A_2\setminus B_2\ \lor\ x\in B_2\setminus A_2)$$

$$i.e.\ (x\in A_1\oplus B_1)\lor(x\in A_2\oplus B_2)$$ $$i.e.\ x\in (A_1\oplus B_1)\cup (A_2\oplus B_2)$$

therefore $(A_1\cup A_2)\oplus (B_1\cup B_2)\subset (A_1\oplus B_1)\cup (A_2\oplus B_2)$.

There's also another way using this result: $\ x\in A\oplus B\ iff\ 1_A (x)\neq 1_B (x)\ $but don'y know how to aprroach with this one. Any ideas?


Yes, it seems correct.

  1. It can also be done without referring to elements, using only the set operations and their properties. $\def\1{{\bf1}}$
  2. $x\in (A_1\cup A_2)\oplus(B_1\cup B_2)\ \iff\ \1_{A_1\cup A_2}(x)\ne \1_{B_1\cup B_2}(x)$.
    Now use that $\1_{A_1\cup A_2}(x)=\max(\1_{A_1}(x),\,\1_{A_2}(x))$ to conclude that at least one of the inequalities $\1_{A_i}(x)\ne \1_{B_i}(x)$ hold ($i=1,2$), hence $x\in (A_1\oplus B_1)\cup (A_2\oplus B_2)$.
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