made easy test series

Question

Consider P and Q be language over Σ = {0, 1} represented by the regular expression 0* (10*)* and (0* + 1*)* respectively. Which of the following is true?

A) P⊂Q

B) Q⊂P

C) P=Q

D)P∩Q=0*1*

Comments

incomplete question!

Answer 1

( 0* + 1* ) * is equivalent to (0+1)* = ∑* ( if you didn't get why these two are equivalent, then think what can't we form with ( 0* + 1* ) * )

( 1 . 0* ) * is equivalent to ( 1 . ( 0 + 1 )* + ∈ )= All Strings Starts with 1 + empty String

0* ( 1 . 0* ) * is equivalent to  0* . ( 1 . (0+1)* + ∈ ) = (0+1)* = ∑*

 

therefore P=Q

Comments

@talha hashim, thanks for correcting me

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( 1 . 0* ) * is equivalent to  1 . ( 0 + 1 )* = All Strings Starts with 1 + empty String

I have doubt on the above statement.
The (1.0*)* can generate Null whether 1.(0+1)* can't.

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Typing mistake bro..... we have to add ∈ there.... thanks for that