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There are three propositions A, B, C and another condition "some two of these propositions are true and the third one is false", or, in other words, "exactly 2 of 3 propositions are true". Using truth tables and a Karnaugh map (as discussed at How to find the logical formula for a given truth table?) i deducted the Boolean expression for this: ABC' + AB'C + A'BC. Is there any more succinct notation for this expression in any branch of logic?

Edit: Obviously using proposition calculus notation the above statement may be represented as: $(A \wedge B \wedge \neg C) \vee (A \wedge \neg B \wedge C) \vee (\neg A \wedge B \wedge C)$. I am sorry if that misguided you. I'm still interested, if any more succinct notation is possible.

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    $\begingroup$ I have never seen such a notation. Don't forget, if you are writing up something where you need this a lot, you can always invent ad hoc notation of your own. Just don't forget to explain it, and also beware that too much ad hoc notation may hinder rather than help communication, especially if it's not well thought out. $\endgroup$ Aug 8, 2012 at 11:22
  • $\begingroup$ @HaraldHanche-Olsen I think the notation is similar to the boolean algebra operation of $\cdot$ and $+$ with the $\cdot$ implicitily used between letters. $\endgroup$
    – William
    Aug 8, 2012 at 14:00
  • $\begingroup$ If you are using the symbol for your own purposes (taking notes, studying, etc.), then invent any symbol you like. If you are planning to use this symbol on a manuscript you are expecting anyone else to read, then I strongly suggest you do not use such a symbol. I personally believe that one should do everything possible to make their own papers as easy to read as possible. In my opinion, if you are not willing to put in the work to make your paper as easy to digest as possible, then why should others put in the work to read the paper? I would highly recommend simply using words. $\endgroup$
    – JavaMan
    Aug 8, 2012 at 20:23

2 Answers 2

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Using the Iverson bracket, $$ [A]+[B]+[C]=2 $$

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The most symmetric definition of 'exactly one of three' I know is $$ \text{exactly one of } P, Q, R \text{ is true} \;\equiv\; (P \equiv Q \equiv R) \land \lnot (P \land Q \land R) $$ This uses the fact that equivalence ($\;\equiv\;$) is associative.

Using this, we can write \begin{align} & \text{exactly two of } A, B, C \text{ are true} \\ = & \;\;\;\;\;\text{"invert the count"} \\ & \text{exactly one of } A, B, C \text{ is false} \\ = & \;\;\;\;\;\text{"$\;P \equiv \text{false}\;$ is the same as $\;\lnot P \equiv \text{true}\;$ (three times)"} \\ & \text{exactly one of } \lnot A, \lnot B, \lnot C \text{ is true} \\ = & \;\;\;\;\;\text{"the above definition"} \\ & (\lnot A \equiv \lnot B \equiv \lnot C) \land \lnot(\lnot A \land \lnot B \land \lnot C) \\ = & \;\;\;\;\;\text{"simplify"} \\ & \lnot(A \equiv B \equiv C) \land (A \lor B \lor C) \\ \end{align}

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  • $\begingroup$ The double use of $\equiv$ as a binary and ternary operatior is hideous and confusing. $\endgroup$
    – Lord_Farin
    Nov 18, 2013 at 18:14
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    $\begingroup$ @Lord_Farin Obviously we have different tastes at this point. :-) If $\;(P \equiv Q) \equiv R\;$ is equivalent to $\;P \equiv (Q \equiv R)\;$, as it is in classical logic, why write the parentheses? Also, the associativity and symmetry of $\;\equiv\;$ make a formula like $$P \land Q \;\equiv\; P \lor Q \;\equiv\; P \;\equiv\; Q$$ a very useful tool in logical calculations. As to confusion: that depends on what you are used to. If that downvote is yours: I don't really see how this answer deserves it. It is correct and it answers the question... $\endgroup$ Nov 18, 2013 at 18:19
  • $\begingroup$ @Lord_Farin I changed the calculation to use $\;=\;$ throughout. That is actually what Dijkstra/Scholten and Gries/Schneider do in their work (EWD1300 and A Logical Approach to Discrete Math), if I recall correctly. $\endgroup$ Nov 18, 2013 at 18:23
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    $\begingroup$ Indeed, I disagree with that practice. The notation is used commonly in two different ways (yours, apparently, and mine). That makes it a source of confusion, and I would never recommend someone to use such a thing. I'm not saying you didn't give an answer, but I consider it not to be a useful one. My downvote firmly stands. $\endgroup$
    – Lord_Farin
    Nov 18, 2013 at 18:34
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    $\begingroup$ +1 for firmness, although obviously I disagree: if one uses logic as a tool (as opposed to an object of study) then this is one of the things that is actually helpful. Therefore I prefer the other way to avoid confusion, which is to never write something like $\;\ldots = \ldots = \ldots\;$, and instead always use the calculational multi-line format as I used in this answer. $\endgroup$ Nov 18, 2013 at 18:39

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