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Deductive agruments in finite math: wrong answer in book?

  1. Aug 10, 2017 #1
    1. The problem statement, all variables and given/known data

    Identify the following as a valid or an invalid argument.

    pq
    qr
    --------------

    ∴ ~r → ~p

    2. Relevant equations

    N/A

    3. The attempt at a solution

    Truth table values:

    (a) pq TTFFTTTT
    (b) qr TFFFTFFF
    (c) a ∧ b TFFFTFFF
    (d) ~r → ~p TFTFTTTT
    (e) c → d TTTTTTTT

    Since (e) is true in every case, the argument is valid. However, the answer in the back of the book says invalid.

    Am I doing something wrong, or is the answer in the book wrong?
     
  2. jcsd
  3. Aug 10, 2017 #2

    haruspex

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    I get your answer, but just from reasoning. You are given r and q are true, so the predicate not r leads to contradiction. A false predicate implies anything at all.
     
  4. Aug 17, 2017 at 7:11 PM #3

    Mark44

    Staff: Mentor

    It's hard to tell what you mean by this and the other logical expressions you show.
    For a truth table for ##p \Rightarrow q##, you need three columns: one for p, one for q, and one for the implication, like this:
    ##\begin{array} {ccc} \\ p & q & p \Rightarrow q \\
    T & T & T \\
    T & F & F \\
    F & T & T \\
    F & F & T \end{array}##
    The only pair of values for p and q for which the implication is false is when p is true and q is false. In all other cases, the implication is true.

    Edit: Miswrote something in my truth table, but now fixed.
     
  5. Aug 17, 2017 at 7:14 PM #4

    haruspex

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    Shouldn't the last line in the table read FFT?
     
  6. Aug 17, 2017 at 7:17 PM #5

    Mark44

    Staff: Mentor

    Yes -- now fixed.
     
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