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Show that the conditional statement is a Tautology without using truth tables

  1. Sep 10, 2007 #1
    1. The problem statement, all variables and given/known data

    Show that [tex]\left[\neg\,p\,\wedge\,\left(p\,\vee\,q\right)\right]\,\longrightarrow\,q[/tex] is a tautology without using truth tables.



    2. Relevant equations

    DeMorgan's Laws, etc.



    3. The attempt at a solution

    [tex]\left[\neg\,p\,\wedge\,\left(p\,\vee\,q\right)\right]\,\longrightarrow\,q[/tex]

    by. EX 3 (see EX 8)

    [tex]\left[\neg\,p\,\wedge\,\left(p\,\vee\,q\right)\right]\,\vee\,q[/tex]

    [tex]\left[p\,\wedge\,\neg\,\left(p\,\vee\,q\right)\right]\,\vee\,q[/tex]

    [tex]\left[p\,\wedge\,\left(\neg\,p\,\wedge\,\neg\,q\right)\right]\,\vee\,q[/tex]

    [tex]\left[\left(p\,\wedge\,\neg\,p\right)\,\wedge\,\left(p\,\wedge\,\neg\,q\right)\right]\,\vee\,q[/tex]

    [tex]\left[F\,\wedge\,\left(p\,\wedge\,\neg\,q\right)\right]\,\vee\,q[/tex]

    Now what?
     
  2. jcsd
  3. Sep 10, 2007 #2
    there is an error in your first line

    a -> b is logicaly equivlent to ~a or b
     
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