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1. {$wxw$ | x,w belong to $(0+1)^*$}    = = $(0+1)^*$

2. {$wxw$ | x,w belong to $(0+1)^+$}   = = $0(0+1)^+0 + 1(0+1)^+1$

3. {$wxw^r$ |x,w belong to $(0+1)^+$}  = = $0(0+1)^+0 + 1(0+1)^+1$

4. {$xww^rx$ | x,w belong to $(0+1)^+$}  = = $(0+1)^+00(0+1)^+ + (0+1)^+11(0+1)^+$

5. {$xwyw$ | x,w belong to $(0+1)^+$}  = = $(0+1)^+0(0+1)^+0 + (0+1)^+1(0+1)^+1$

1 Answer

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1. Correct. If we put epsilon in place of w, we will get L=(0+1)*.

2. Incorrect.01101 should be involved in the language, which is not involved in the regular expression in right.

3.Correct.Regular expression totally covers all possibilties of the language.

4.Incorrect. 010011 is accepted by the regular expression which is not a part of the language.

// I think you wanted to write y at the last place instead of x, if that is so, it will be correct.

5.Correct.Regular expression totally covers all possibilties of the language.

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Show that the following pairs of regular expressions define the same language over the alphabet I = [a, b].s(a) p(pp)*( A + p)q + q and p*q(b) A +0(0+1)* + (0+1)* 00(0+1)...