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Homework Help: Simple Logic Truth Table Needs Checking

  1. Sep 13, 2009 #1
    Dear All,

    Having trouble with a seemingly simple logic truth table. Are these answers correct?

    [tex]\begin{tabular}{| c | c | c | c | c | c |}
    p & q & r & (p \vee q)\wedge(q \vee r) & (\neg p \wedge q) \vee ( p \wedge \neg r) & p \rightarrow q \rightarrow r\\
    T & T & T & T & F & T \\
    T & T & F & T & T & F \\
    T & F & T & T & F & T \\
    T & F & F & F & T & F \\
    F & T & T & T & T & T \\
    F & T & F & T & T & F \\
    F & F & T & F & F & T \\
    F & F & F & F & F & T \\
  2. jcsd
  3. Sep 13, 2009 #2


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    Homework Helper

    Hi Hotsuma, I think the and/or/not parts are ok, but for the last part I think it would be

    \begin{tabular}{| c | c | c | c }
    p & q & r & p \rightarrow q \rightarrow r\\
    T & T & T & T \\
    T & T & F & F \\
    T & F & T & F \\
    T & F & F & F \\
    F & T & T & T \\
    F & T & F & T \\
    F & F & T & T \\
    F & F & F & T \\

    This effectively has 2 tests, it first requires that if p then q, if this is satisfied the the further condition that if q then r must be satisfied

    In the 3rd case, as p is T and q is F, then the line is FALSE as the first condition is not satisfied

    For the last 4 cases as p is F then the line is T by default, and we don't go any further - its a vacuous truth

    what do you think?
    Last edited: Sep 13, 2009
  4. Sep 13, 2009 #3
    Dear lanedance,

    I considered this but wasn't sure if that was correct or not. What you described to me in your past post could be written as "if p then q" and "if q then r," which I don't think is the same thing, or rather, I'm not sure whether it is or is not. Thanks for the suggestion though, having someone else recommend what I was thinking is always appreciated.
  5. Sep 13, 2009 #4
    And I'd love to prove that the two are tautologies, but since I am not sure about the former, that doesn't really help me much :(.

    [tex](p \rightarrow q) \wedge (q \rightarrow r) \equiv p \rightarrow q \rightarrow r [/tex]
  6. Sep 13, 2009 #5


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    Homework Helper

    I don't think is is exactly the same thing, take the case:
    p = F
    q = T
    r = F

    this gives the following:

    [tex] p \rightarrow q [/tex]
    True - vacuously as P is False

    [tex] q \rightarrow r [/tex]

    [tex] (p \rightarrow q) \wedge (q \rightarrow r) [/tex]

    [tex] p \rightarrow q \rightarrow r [/tex]
    True - vacuously as P is False
    Last edited: Sep 13, 2009
  7. Sep 13, 2009 #6
    There is a problem here. Implication is not an associative connective. i.e.

    [tex](p \rightarrow q) \rightarrow r \not\equiv p \rightarrow (q \rightarrow r)[/tex]

    so the meaning of [itex]p \rightarrow q \rightarrow r[/itex] is ambiguous.

    (Consider the case where p = F, q = T, r = F.)

    Unless there is an assumption about the order of execution of multiple implications, the last column could be evaluated differently (either TFTTTFTF or TFTTTTTT).

  8. Sep 14, 2009 #7
    Yeah, that was my fear. I e-mailed the professor at like 4PM but he has not yet responded (nor do I expect him to).
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