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An Introduction to Formal Languages and Automata, 6th Edition by Linz

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ANSWERS: SOLUTIONS AND HINTS FOR SELECTED EXERCISES

Chapter 1

Section 1.1

5. Suppose xST. Then xS and xT, which means xS and xT¯, that is xST¯. So STST¯. Conversely, if xST¯, then xS and xT, which means xST, that is ST¯ST. Therefore ST = ST¯.

6. For Equation (1.2), suppose xS1S2¯. Then xS1S2, which means that x cannot be in S1 or in S2, that is xS¯1S¯2. So S1S2¯S¯1S¯2. Conversely, if xS¯1S¯2, then x is not in S1 and x is not in S2, that is, xS1S2¯. So S¯1S¯2S1S2¯. Therefore S1S2¯ = S¯1S¯2.

12. (a) reflexivity: Since |S1| = |S1|, so reflexivity holds.

(b) symmetry: if |S1| = |S2| then|S2| = |S1| so symmetry is satisfied.

(c) transitivity: Since ...

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