symmetric difference of languages - both are in NP and coNP

Question

I have this problem,

Let $L_1,L_2$ be languages in $NP \cap co-NP$. I want to show that their symmetric difference is also in $NP \cap co-NP$. Like:

$L_1 \oplus L_2$ = {x | x is in exactly one of $L_1, L_2$}

I do not have a clue how to show it. We know that $L_1 \cap L_2 \in NP$ is unknown. So for that reason it is reasonable to ask only that instance of the problem. From my point of view, if $L_1 \in NP$ there is some verifier for that language which runs in polynomial time. We have such verifier for the second language $L_2$ too. My proposal of the machine M which decides $L_1 \oplus L_2$ is as follows:

Let's have

M = "for the input x:
1. copy x on the second tape
2. run x on M_1 on the first tape
3. if M_1 accepts, (otherwise go to 4.)
    3a) run x on M_2
    3b) if  M_2 accepts, M rejects 
4. run x on M_2, if M_2 accepts, M accepts, otherwise M rejects.

I do not know the relation of this with the co-NP class ... Is my reasoning right? This machine works like a charm for languages $L_1,L_2 \in P$. Does it hold also for that intersection?

Thank you a lot

Answer 1

Yes, the class NP $\cap$ coNP is closed under symmetric difference. To see this, suppose that $A$ and $B$ are both in NP $\cap$ coNP. This means that the truth of $a\in A$ can be verified in polynomial time with a suitable witness, and also $a\notin A$ can be verified in polynomial time with a suitable witness, and the same for $B$. Let $A\triangle B=(A-B)\cup (B-A)$ be the symmetric difference (the same as what you call $A\oplus B$), and argue as follows:

• $A\mathrel{\triangle} B$ is in NP. This is because we can verify whether $a\in A\mathrel{\triangle} B$ by verifying either that $a\in A$ and $a\notin B$ or that $a\notin A$ and $a\in B$. That is, the objects $a$ in the symmetric difference $A\mathrel{\triangle} B$ are verified by a pair of witnesses, which either verify membership in $A$ and not in $B$ or in $B$ but not in $A$.

• $A\mathrel{\triangle} B$ is in coNP. This is because we can verfiy whether $a\notin A\mathrel{\triangle} B$ by verifying either that $a\in A$ and $a\in B$ or that $a\notin A$ and $a\notin B$.

So the symmetric difference is in NP $\cap$ coNP, as desired.

(Meanwhile, your proposed algorithm, if interpreted as a nondeterminisitic algorithm, is not correct, since failure to verify membership nondeterministically is not the same thing as a verfication of non-membership. In short, your algorithm can be fooled in the following way: suppose $x$ is in both languages, but in step 2 of your algorithm, $M_1$ happens to choose a branch of computation that doesn't verify membership---an inadequate witness, so it doesn't accept on that branch---but then $M_2$ does accept $x$ in step 4. In this case, you will accept $x$ when you shouldn't.)

Answer 2

NP is closed under (finite) unions and intersections, therefore using De Morgan laws, $\mathrm{NP}\cap\mathrm{coNP}$ is closed under unions and intersections. Since it is also trivially closed under complement, it is closed under all Boolean operations, including the symmetric difference.

A more careful argument shows that $\mathrm{NP}\cap\mathrm{coNP}$ is even closed under polynomial-time Turing reductions. (Note that symmetric difference can be computed by an essentially constant-time algorithm querying oracles for the two languages.) In fact, $(\mathrm{NP}\cap\mathrm{coNP})^{\mathrm{NP}\cap\mathrm{coNP}}=\mathrm{NP}\cap\mathrm{coNP}$.