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Proof by contradiction for statement of the form P->(Q and R)

  1. Jul 18, 2012 #1
    Say I have a statement like this:
    P implies (Q1 and Q2).

    If I wanted to prove this by contradiction, I would assume P and not(Q1 and Q2)=[(not Q1) or (not Q2)] both hold, and try to find a contradiction.

    My question is... Am I done if I find a contradiction while assuming P and [(not Q1) and (not Q2)] ? Is this sufficient? Or do I need to find a contradiction in both the statements:
    P and (not Q1),
    P and (not Q2)

  2. jcsd
  3. Jul 18, 2012 #2
    Never mind. I figured it out.
  4. Jul 19, 2012 #3


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    No. The negation of "Q and R" is "not Q or not R".
    The negation of "if P then (Q and R)" is "If (not Q or not R) then not P".

    (For those who read this thread and wondered).
  5. Jul 19, 2012 #4


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    The contrapositive of "if P then (Q and R)" is "if (not Q or not R) then not P"

    But the contrapositive has the same truth value as the original.

    The negation of "if P then (Q and R)" is "P and (not Q or not R)"
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