I am trying to understand what the operator $*$ means in Boolean algebra. If x is a Boolean variable, what does the expression $x*$ mean?
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1Is it $x$, or $xy$? And are you sure that the symbol looks like $*$, or are you talking about multiplication? – Arthur Sep 07 '19 at 20:35
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It is $x$ not $xy$. That is, it is a unary operator. It is from the book, "Discrete Mathematics and its Application" by Kenneth Rosen. It is used in one of the exercises. I could believe that it is a typo but I think not. – Bob Sep 07 '19 at 20:40
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2Well, if it's used in an exercise, the symbol will have been introduced somewhere. – egreg Sep 07 '19 at 20:54
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I have that book but do not find the $ x*$ expression in the chapter dealing with Boolean algebra. Could you specify the page? – Anatoly Sep 07 '19 at 21:37
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Look on page 39 which is section 1.3 There are exercises for the reader to do. Look at exercise numbers. Look at exercises 40 and 41. You will see $s*$ used where s is a Boolean variable / compound proposition. – Bob Sep 07 '19 at 22:33
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1You might want to include a screen shot or update the question with the specific text. – Axion004 Sep 08 '19 at 02:21
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1Could it be that the author uses $$ for the complement, as $x^$? I've seen that before. – amrsa Sep 08 '19 at 10:36
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@amrsa It could be the complement. I am not sure. – Bob Sep 09 '19 at 12:50
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In these exercises, $s*$ means the dual of $s.$ This is obtained from $s$ (which is an expression of propositional calculus containing only the operations $\wedge, \vee,$ and $\neg$) by replacing $\wedge$ with $\vee$, $\vee$ with $\wedge$, $T$ with $F$, and $F$ with $T$.
In the edition of the book I looked at, the author defines dual just before exercise 34, a few exercises above the one the OP referenced in a comment.
Mitchell Spector
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