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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

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1 Answer 1

We know $L_1,L_2,\overline{L_1},\overline{L_2} \in NP$. To show that the symmetric difference of $L_1$ and $L_2$ lies in $NP \cap coNP$, we need to find two verifiers, one for $L_1 \otimes L_2$ and one for $\overline{L_1 \otimes L_2}$. A certificate that some input $x$ is in $L_1 \otimes L_2$ must guarantee that $x \in L_2, x \notin L_2$ or $x \in L_2, x \notin L_1$. So a machine to verify such a certificate can be built out of the verifiers we know exist for $L_1,L_2$ and their complements. A similar construction will work for $\overline{L_1 \otimes L_2}$.

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Isn't there a typo? you wrote $x \in L_2, x \notin L_2$, shouldn't be that first $L_2$ $L_1$ instead? –  Stefan.M Mar 21 '12 at 0:34
    
If I have some certificate $w$ for which the verifier V verifies in polynomial time if $x$ is in $L_1$, then if the answer is true, I put this certificate on the verifier for the L_2 and this has to be false? –  Stefan.M Mar 21 '12 at 0:41
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