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Truth statement/Truth table

  1. Sep 6, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that the statement below is neither a tautology or a contradiction
    [ p V ((¬ r) → (¬s))] V [ ( s → (( ¬ t ) V p )) V ((¬ q ) → r )]

    2. Relevant equations



    3. The attempt at a solution

    My thing is there way to do this...I don't know if this makes sense but i think it is neither because they have 2 variables that don't exist a regular Q and a regular T....I am somewhat new at this and I think because there isn't regular Q and T I think this statement is neither a tautology or contradiction because the statement has a ~Q and a ~T. PLEASE HELP....
     
  2. jcsd
  3. Sep 8, 2013 #2

    NascentOxygen

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    Staff: Mentor

    Hi ARTIE24M. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    I don't know how to answer this question, either. But I don't think you can write it off simply on the basis that the second term includes 2 variables that don't appear in the first. (After all, it might be that the ~q is effectively ANDED with zero, therefore removing variable q from the expression.)

    The way I'm thinking you might be expected to solve it is to replace each implication A ⟶ B with the logical equivalent ¬A V B until you have simplified the whole expression to something like: p V q V r V s V ¬t

    and you can see this is neither always TRUE, nor always FALSE.

    If you find out your examiner requires something completely different, please post here.

    Reference: http://www.millersville.edu/~bikenaga/math-proof/truth-tables/truth-tables.html
     
    Last edited by a moderator: May 6, 2017
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