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Proof by contradiction

  1. Apr 10, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose I want to prove the following statement by contradiction:
    [itex] P \longrightarrow (Q \land Z) [/itex]

    2. Relevant equations

    If [itex] (Q \land Z) [/itex] is false, then either: (i) Q is false and Z is true; (ii) Q is true and Z is false; (iii) Q and Z are false.

    3. The attempt at a solution

    Do I need to consider all possible cases in which [itex] (Q \land Z) [/itex] is false and arrive to a contradiction or it suffices to show a contradiction in only one possible of the three possible cases?

    Thanks for your help!
     
  2. jcsd
  3. Apr 10, 2012 #2
    If you want to prove the first one is true all you have to do is prove it by only one of the cases, because QandZ is false only if one of those cases occurs, but depending on what your teacher wants you can do all three cases.
     
  4. Apr 10, 2012 #3
    This means that the proof is complete if I assume (say) case (i) to be true and arrive to a contradiction?
     
  5. Apr 10, 2012 #4
    Yes because Q and Z could only be true if Q and Z are both true or if Q and Z are both false, therefore all you have to do is prove it using one case. Hope this helps you.
     
  6. Apr 10, 2012 #5
    Think I got it. Thanks a lot!
     
  7. Apr 10, 2012 #6
    No Problem.
     
  8. Apr 13, 2012 #7
    Hi again,

    It appears that your answer is not completely correct or I am truly messed up. Negating (Q and Z) means that either (Q and not Z) or (Z and not Q). Hence, to actually show that P implies (Q and Z), don't we need to show that both of the above cases aren't possible true? I mean, showing that P implies (Z and not Q) does not exclude the possibility that (Q and not Z) still is true, does it?

    Thanks!!!!
     
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