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Homework Help: Logical equivalence of statements with truth tables

  1. Dec 31, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex](S \wedge T) \wedge \neg (S \wedge T)[/tex]

    [tex] (S \vee T) \Rightarrow (S \wedge T) [/tex]

    Are the two (predicates?) logically equivalent?
    2. Relevant equations

    Not sure, but I believe that logical equivalence means that two predicates give the same output on the same input.

    3. The attempt at a solution

    Worked truth tables. The end result of both where statement T is t t f f and statement S is t f t f, in both cases is t f f t. I guess I am not really sure if that means that the two predicates are logically equivalent or not. I was actually just playing with truth tables and got this on accident, and wasn't sure if the fact that they both have the same end result is worth noting or not.
  2. jcsd
  3. Dec 31, 2008 #2


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    I will take a look at this in the afternoon if other PF'ers haven't got to it before me...its 2.30 in the morning.
  4. Dec 31, 2008 #3
    The way I see it is you just work the logic tables look at all possibilities for (S,T) i.e. (True, True), (True, False), (False, True), (False, False). If that's what you did and got the same result then I would believe they are logically equivalent.
  5. Dec 31, 2008 #4


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    Yes, that is correct. If this is homework, surely you could look that up in your textbook?

    You just said " logical equivalence means that two predicates give the same output on the same input"
  6. Dec 31, 2008 #5


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    Homework Helper

    First, note that

    (S \wedge T) \wedge \neg (S \wedge T)

    will always be false (for every combination of S and T) since [tex] (S \wedge T) [/tex] and [tex] \neg (S \wedge T) [/tex] always have opposite truth values.

    However, if both S and T are true, then so is

    (S \vee T) \Rightarrow (S \wedge T)

    These two statements are not equivalent.
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