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Math Help - Rewriting expressions with logic laws (4)

  1. #1
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    Rewriting expressions with logic laws (4)

    Rewrite: q \wedge ( \sim (((p \vee (p \wedge q)) \wedge (p \vee q)) \vee q)) and determine if its tautology, contradiction or neither. Using truth table I got contradiction.

    My working out:

    q \wedge ( \sim (((p \vee (p \wedge q)) \wedge (p \vee q)) \vee q))
    \equiv q \wedge ( \sim ((p \wedge (p \vee q)) \vee q))
    \equiv q \wedge ( \sim (p \vee q))
    \equiv q \wedge ( \sim p \wedge \sim q)
    \equiv q \wedge ( \sim q \wedge \sim p)
    \equiv (q \wedge \sim q) \wedge \sim p
    \equiv p \wedge \sim p
    \equiv F

    So I finished off with a contradiction. Is this correct? Any help would be much appreciated.
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  2. #2
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    Can anyone confirm? thanks
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  3. #3
    Super Member Failure's Avatar
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    Quote Originally Posted by jvignacio View Post
    Rewrite: q \wedge ( \sim (((p \vee (p \wedge q)) \wedge (p \vee q)) \vee q)) and determine if its tautology, contradiction or neither. Using truth table I got contradiction.

    My working out:

    q \wedge ( \sim (((p \vee (p \wedge q)) \wedge (p \vee q)) \vee q))
    \equiv q \wedge ( \sim ((p \wedge (p \vee q)) \vee q))
    \equiv q \wedge ( \sim (p \vee q))
    \equiv q \wedge ( \sim p \wedge \sim q)
    \equiv q \wedge ( \sim q \wedge \sim p)
    \equiv {\color{red}(q \wedge \sim q)} \wedge \sim p
    So far so good. But now things get slightly off the rails:
    \equiv {\color{red}p} \wedge \sim p
    \equiv F
    This should, imho, be
    \equiv {\color{blue}F}\wedge \sim p

    \equiv F
    So I finished off with a contradiction. Is this correct?
    The result is correct, in my opinion, but see the above glitch in the derivation.
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  4. #4
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    Quote Originally Posted by Failure View Post
    This should, imho, be
    \equiv {\color{blue}F}\wedge \sim p
    Quick question is {\color{blue}F}\wedge \sim p the same as \sim p \wedge {\color{blue}F} ? and they both = {\color{blue}F}. According to the Commutative law they should be right?
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  5. #5
    Super Member Failure's Avatar
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    Quote Originally Posted by jvignacio View Post
    Quick question is {\color{blue}F}\wedge \sim p the same as \sim p \wedge {\color{blue}F} ? and they both = {\color{blue}F}. According to the Commutative law they should be right?
    Right, but to be honest, I personally use a much more general rule: whenever there is a "false" anywhere in no matter how large a conjunction ( \wedge), the whole conjunction collapses to the value "false".

    It's akin to having a 0 in a product of numbers: the whole product collapses to the value 0.
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