De Morguns Law | De Morguns Law Proof In Toa

#DeMorgunsLaw
#DeMorgunsLaw Proof
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Statement of de morguns law
If L1 and L2 are two regular languages, then L1 ∩ L2 is also regular.
Proof
Using De-Morgan's law for sets
(L1
c ∪ L2
c
)
c = (L1
c
)
c ∩ (L2
c
)
c = L1 ∩ L2
Since L1 and L2
are regular languages, so are L1
c
and L2
c
. L1
c
and L2
c
being regular provide that L1
c ∪ L2
c
is also
regular language and so (L1
c ∪ L2
c
)
c = L1 ∩ L2
, being complement of regular language is regular language.
Following is a remark
Remark
If L1 and L2 are regular languages, then these can be expressed by the corresponding FAs. Finding regular
expressions defining the language L1 ∩ L2 is not so easy and building corresponding FA is rather harder.
Following are example of finding an FA accepting the intersection of two regular languages

[@amirshaheen5834]

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[@amirshaheen5834]

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