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Need help with truth table for P->Q and it's inverse

  1. Mar 3, 2013 #1
    Hi!

    I'm struggling with the bellow truth tables:

    P = I live in Paris
    Q = I live in France

    A B C
    P → Q Q → P ¬ Q → ¬ P
    S S S S S S F S F
    S F F F S S S F F
    F S S S F F F S S
    F S F F S F S S S

    Table A and C I'm fully clear with. However I don't understand the truth table for B (the converse of P → Q).

    Table B:

    Q → P
    S S S
    F S S
    S F F
    F S F

    If I live in France there is a possibility I live in Paris S (Understand)
    If I don't live in France there is a possibility I live in Paris S (Don't understand)
    If I live in France there is not a possibility I live in Paris F (Don't understand)
    I I don't live in France there is not a possibility I live in Paris S (Understand)

    I want line 2 to be false and line 3 to be true.

    Can someone explain why this should not be the case?
     
    Last edited: Mar 3, 2013
  2. jcsd
  3. Mar 3, 2013 #2

    mfb

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    Q → P ("If I live in France, I live in Paris") is wrong in reality, why do you expect correct results for this?
     
  4. Mar 3, 2013 #3

    micromass

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    If Q is false, then [itex]Q\rightarrow P[/itex] is always a true statement. So if you don't live in France, then the statement "If I live in France, then I live in Paris" is true.

    The only way [itex]Q\rightarrow P[/itex] can ever be false is if [itex]Q[/itex] is true and [itex]P[/itex] is false.
     
  5. Mar 3, 2013 #4

    mfb

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    Okay, to be more precise: Q → P is not true in general (=for all) in reality.
     
  6. Mar 4, 2013 #5

    Bacle2

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    The truth table for Q→P is exactly the same as that for P→Q , since they are both

    plain/standard conditionals. Notice that the tables are exactly the same except

    rows 2 and 3 have been exchanged.
     
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