patternMinor
Two languages such that $L_1 \cup L_2 \leq_m\, L_1 \cap L_2$ and two (other?) such that $L_1 \cap L_2 ≤_m\, L_1 \cup L_2$?
Viewed 0 times
suchl_1languagesleq_mcapcuptwothatl_2and
Problem
Are there languages $L_1$, $L_2$ such that such that
$$L_1 \cup L_2\leq_m\, L_1\cap L_2,$$
and two other languages such that
$$L_1 \cap L_2 \leq_m\, L_1 \cup L_2?$$
And if so, what are they? How can i go about finding these?
$$L_1 \cup L_2\leq_m\, L_1\cap L_2,$$
and two other languages such that
$$L_1 \cap L_2 \leq_m\, L_1 \cup L_2?$$
And if so, what are they? How can i go about finding these?
Solution
How about taking $L_1=L_2$? Then we will have both
$L_1 \cup L_2\leq_m\, L_1\cap L_2$ and $L_1 \cap L_2 \leq_m\, L_1 \cup L_2$.
If you want to have $L_1\not= L_2$, you just try $L_2=L_1\cup\{w\}$, where $w$ is a string not in $L_1$.
If you are determined to have infinity elements in both $L_1\setminus L_2$ and $L_2\setminus L_1$, try $L_1=L\cup aL$ and $L_2=L\cup bL$, where $L$ is an infinite language over $\Sigma$ and $a,b$ are two new symbols not in $\Sigma$.
By now, you should be able to see that there are lots of way to construct examples that satisfy the requirements.
$L_1 \cup L_2\leq_m\, L_1\cap L_2$ and $L_1 \cap L_2 \leq_m\, L_1 \cup L_2$.
If you want to have $L_1\not= L_2$, you just try $L_2=L_1\cup\{w\}$, where $w$ is a string not in $L_1$.
If you are determined to have infinity elements in both $L_1\setminus L_2$ and $L_2\setminus L_1$, try $L_1=L\cup aL$ and $L_2=L\cup bL$, where $L$ is an infinite language over $\Sigma$ and $a,b$ are two new symbols not in $\Sigma$.
By now, you should be able to see that there are lots of way to construct examples that satisfy the requirements.
Context
StackExchange Computer Science Q#110433, answer score: 3
Revisions (0)
No revisions yet.