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Math Help - (p <--> q) and not q - Question

  1. #1
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    (p <--> q) and not q - Question

    Hello,

    I just have a quick question about the statement in the title.

    (p <-> q) AND NOT q

    If I was to find the truth statement from that.

    How am I to break the (p <-> q) down into a (p AND q) type statement first?

    because I assume that if I can get the q out of there, then q AND NOT q together will = False.

    And then any AND statement with a false in it is inevitably False.

    Thanks
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  2. #2
    Moo
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    Hello,

    \wedge=and
    \vee=or
    \bar{a}=non \ a

    p \leftrightarrow q is \{p \rightarrow q\} \wedge \{q \rightarrow p\}

    \{p \rightarrow q\} is \bar{p} \vee q
    \{q \rightarrow p\} is \bar{q} \vee p


    So p \leftrightarrow q is \{\bar{p} \vee q\} \wedge \{\bar{q} \vee p\}


    If you want the formula of your proposition, develop, according to these rules :

    (a \wedge b) \vee c=(a \vee c) \wedge (b \vee c)

    and (a \vee b) \wedge c=(a \wedge c) \vee (b \wedge c)

    and the associativity of \vee and \wedge

    ------------------

    Your reasoning with "necessarily false" is not correct, because it can be "or" instead of "and".
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  3. #3
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    Thanks Moo, I see what I missed now ! I was looking at the answer the whole time. I should have been able to figure out that to break it up it looked like that !

    Thanks for the reasoning part as well

    I think I might have to hit the books again and get a bit more practise under the belt !

    Cheers
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