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Truth Functional (TF): Has a true/false value which can be completely determined by the truthfulness/falsefullness (?) of the input's values (got that?).

Question: Show that "It is likely that __" is not a truth functional operator (where the blank, the "input", is filled with a simple sentence with a true/false value).

We can prove "likely" is not TF by finding a single counterexample. How do we find a counterexample? For any simple sentence I create which has a true/false value, I already know if it's likely or not.

For example: $A \equiv$ "The Yankees won the world series", is True. It's also likely. Now I need another sentence, $B$, which is True, but is unlikely. Right? Would like explanation and example (or explanation by way of example). Thx.

Edit: P.S. You cannot answer by saying "likely" is a prediction which is not TF. I want to know why "likely" (or even any prediction) is not TF

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If you choose a real number in the interval $[0,1]$ it is likely that this number will be irrational; I chose $1$ which is rational. –  Asaf Karagila Jul 1 '12 at 19:27
    
If you want something more... real life related: it is unlikely that anyone will be born with two faces. However there was this baby in India that was born like that not long ago. –  Asaf Karagila Jul 1 '12 at 19:47
    
What do you mean by likely? That means, what is the semantic meaning of $X \mapsto $ "it is likely that $P(X)$"? –  dtldarek Jul 1 '12 at 19:49
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@dtldarek: "Likely" is not really defined. It's a regular english sentence, which is to be symbolized in logic. There is no specific probability which defines likely in this (I'm really hoping your comment wasn't rhetorical, here). –  Jeff Jul 1 '12 at 21:32
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@Jeff The problem is that depending on the definition of "likely", it may be or might not be a TF. For example, if you define "likely" to depend on probability and you are free to pick your $\omega \in \Omega$, then this is like in Asaf's comment. However, I think that you could define "likely" the way, that only always true things can be likely. You cannot write a sound proof when you have no definition for some of the terms you use. –  dtldarek Jul 1 '12 at 21:57
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1 Answer

up vote 5 down vote accepted

I wouldn't go with finding a concrete counterexample; that seems to get us into a semantic morass that just detracts from the main matter.

Rather, I would say: By definition a "truth-functional operator" is one where the truth of "it is likely that $X$" depends only on whether "$X$" itself is true or not. Thus, if "likely" is a truth functional, then it must have a truth table, and that truth table must explain everything about what "likely" means.

"Likely" has only one argument, so there are only four different possible truth tables it could have:

  X   | X likely       X   | X likely      X   | X likely     X   | X likely
  ----+---------       ----+---------      ----+---------     ----+---------
  Yes | Yes            Yes | No            Yes | Yes          Yes | No
  No  | Yes            No  | Yes           No  | No           No  | No

But clearly neither of these matches any meaning that the word "likely" can have in English:

  • The first one would mean that everything is likely.
  • The second one would mean "it is likely that $X$" means the same as "$X$ is not the case".
  • The third one would mean that "likely" means nothing at all; "it is likely that $X$" would just be a more verbose way to assert $X$ itself.
  • The fourth one would mean that nothing is ever likely.

Since none of the possible truth tables work, "likely" does not have a truth table, and so it is not a truth-functional operator.

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This is easy. The exact question says "Show that .... is not truth funcitonal" - it does not specify how to show it. My only concern is if the prof will accept this type of answer for a similar test question. –  Jeff Jul 2 '12 at 3:02
    
Why wouldn't he? –  Henning Makholm Jul 2 '12 at 12:57
    
On a roll of a die, the odds of any particular side turning up are only one in six. But one always does. So nothing is ever likely. Your fourth table is correct! –  MikeC Jul 2 '12 at 14:06
    
@henning: Well, the book and the prof's class notes only showed counterexamples. –  Jeff Jul 2 '12 at 17:14
    
@FYI: I am told that she got all of the "prove <operator> is not TF" questions correct. Therefore the prof accepted this argument. –  Jeff Jul 3 '12 at 2:46
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