Peter Linz 4th edition Ex-1.2 -Q10

Let take L = { W₁ , W₂, W₃, W₄, ........}

my convention is ( W_i )^R = X_i.

we know that (W₁.W₂ )^R = (X₂ . X₁)

ex:- let W₁ = ab ( this means X₁ = ba ), W₂ = cd ( this means X₂ = dc ) ===> (W₁ . W₂)^R = (ab.cd)R = (dc . ba) = (X₂. X₁)

what is L* ?

L* ={ ∈, L , L², L³,.......} (simply concentrate on L² only )

    = { ∈, ( W₁ , W₂, W₃, W₄, .......) , ( W₁W₁ , W₁W₂, W₁W₃, W₁W₄, .......) , ( W₂W₁ , W₂W₂, W₂W₃, W₂W₄, .......) , .....  }

( L* )^R = simply reverse each string in the L*

  = { ∈, ( (W₁)^R, (W₂)^R, (W₃)^R, .......) , ( ( W₁W₁)^R , (W₁W₂)^R, (W₁W₃)^R, .......), ( ( W₂W₁)^R , (W₂W₂)^R, (W₂W₃)^R,  .......), .....  }

  = { ∈, ( (X₁), (X₂), (X₃), (X₄), ...........) , ( ( X₁X₁) , (X₂X₁), (X₃X₁), .................), ( (X₁X₂) , (X₂X₂), (X₃X₂), , .......), .....  }

what is L^R ?

L^R = reverse each element in the language

   = { (W₁)^R, (W₂)^R, (W₃)^R, ......}

  = { (X₁), (X₂), (X₃), (X₄), ........... }

(L^R)* = {∈, ( (X₁), (X₂), (X₃), (X₄), ...........) , ( ( X₁X₁) , (X₁X₂), (X₁X₃), .................), ( (X₂X₁) , (X₂X₂), (X₂X₃), , .......), .....}

       = { ∈, ( (X₁), (X₂), (X₃), (X₄), ...........) , ( ( X₁X₁) , (X₂X₁), (X₃X₁), .................), ( (X₁X₂) , (X₂X₂), (X₃X₂), , .......), .....  }

       = (L*)^R

NOTE :-

  word X_i . X_j is in ( L^R )^* it must be in ( (L*)^R) converse also.

think why this is true? if you didn't get in general, take the sample strings and realize how it should be true.