Hey Somak Sengupta,
What about this proof ?
(00)*(e+0) = (00)*0 [ e+P = P ]
(00)*0 = 0(00)* [ (PQ)*P = P(QP)* ]
This way (i) & (iv) are equivalent.

Debasish Ray Chawdhuri answered

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Somak Sengupta (anonymous)

(i) and (iii) are equivalent.

Somak Sengupta answered

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Debasish Ray Chawdhuri (anonymous)

i and iv I guess.

Debasish Ray Chawdhuri answered

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Somak Sengupta (anonymous)

No, its (i) and (iii)

(00)*(e+0)= any number of zeros(even+0dd)+ empty string
(00)*= empty string + even number of zeros
0*0*=any number of zeros+empty string
0(00)*=odd n of zeros

Somak Sengupta answered

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Debasish Ray Chawdhuri (anonymous)

Oops,

i think i used one step wrong:
empty string + p = p
erom formula nei……