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Math Help - equivalences help

  1. #1
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    Question equivalences help

    I have to verify the following, using equivalences. Can anyone help me please:


    a) [( p q) and ( p not q)] not p

    b) [(p and q) p ] [p q]

    c) not ( p q) (p not q)

    d) [(p and q r) and (p and not q) not r)] [p (q r)]
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  2. #2
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    a)
    \begin{gathered}<br />
  \left( {p \to q} \right) \wedge \left( {p \to \neg q} \right) \hfill \\<br />
  \left( {\neg p \vee q} \right) \wedge \left( {\neg p \vee \neg q} \right) \hfill \\<br />
  \neg p \wedge \underbrace {\left( {q \vee \neg q} \right)}_{\text{TRUE}} \hfill \\<br />
  \neg p \hfill \\ <br />
\end{gathered}
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  3. #3
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    Can anyone double check my working please? I need to know if I'm doing these things correctly.

    Thanks

    [ ( P Q ) ] ( P Q )
    [ ( P Q ) ) ] [ P ( P Q ) ] If and only if
    [ (P Q ) P ] [ P ( P Q ) ] Implication
    [ ( P Q ) P ] [ P ( P Q ) ] De Morgans
    [ ( P P ) Q ] [ P ( P Q ) ] Associativity
    [ True Q ] [ P ( P Q ) ] Law of Excluded Middle
    [ P ( P Q ) ] Removal Of Constant
    ( P P ) ( P Q ) Distributivity
    P Q Law of Excluded Middle
    P Q Implication
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  4. #4
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    Hello, barc0de!

    (c)\;\;\sim(p \Longleftrightarrow q) \;=\;(p \Longleftrightarrow \:\sim q)

    We have: . \sim\bigg[(p \to q) \wedge (q \to p)\bigg]

    . . . . . . =\;\;\sim\bigg[(\sim p \vee q) \wedge (\sim q \vee p)\bigg]

    . . . . . . = \;\;\sim(\sim p \vee q) \:\vee \sim(\sim q \vee p)

    . . . . . . =\quad(p\: \wedge \sim q)\:\vee (q \:\wedge \sim p)

    . . . . . . = \;\bigg[(p\: \wedge \sim q) \vee q\bigg] \wedge \bigg[(p \:\wedge \sim q) \:\vee \sim p\bigg]

    . . . . . . = \;\bigg[(p \vee q) \wedge (q \:\vee \sim q)\bigg] \wedge \bigg[(p \:\vee \sim p) \wedge (\sim q \:\vee \sim p)\bigg]

    . . . . . . = \;\bigg[(p \vee q) \wedge t\bigg] \wedge \bigg[t \wedge (\sim q \:\vee \sim p)\bigg]

    . . . . . . = \;(p \vee q) \wedge (\sim q \:\vee \sim p)

    . . . . . . = \;(q \vee p) \wedge (\sim p \:\vee \sim q)

    . . . . . . = \;(\sim q \to p) \wedge (p \to \:\sim q)

    . . . . . . = \;\;p \Longleftrightarrow \:\sim q


    Your work on part (b) is correct . . . Nice work!

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  5. #5
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    Thank you very much Soroban for taking the time to check for me. I will post the other two when I have figured them out.

    Again thanks,

    barc0de
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  6. #6
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    Hi Soroban. I don't understand the last bit of your solution. Please could you tell me how you got from this:


    \;(q \vee p) \wedge (\sim p \:\vee \sim q)


    to this:


    <br />
\;(\sim q \to p) \wedge (p \to \:\sim q)<br />

    Thanks
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  7. #7
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    You are expected to know basic facts.
    \begin{gathered}<br />
  a \to b \equiv \neg a \vee b \hfill \\<br />
  a \to b \equiv \neg b \to \neg a \hfill \\<br />
  \neg c \vee \neg d \equiv c \to \neg d \equiv d \to \neg c \hfill \\ <br />
\end{gathered}
    Last edited by mr fantastic; January 10th 2009 at 07:24 PM.
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