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Thread: [SOLVED] Boolean Simplification

  1. #1
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    [SOLVED] Boolean Simplification

    Hi guys,

    I've tried to simplify a formula, and I'm not sure if I've done it right. The following is my solution - I've indicated which logic rule(s) I've used to arrive at each line (excuse the rubbish handwriting):

    http://www.zaraphrax.net/~fagg/boolean_mhf.jpg

    Any guidance is appreciated.

    Thanks!
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  2. #2
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    Quote Originally Posted by Ashgasm View Post
    Hi guys,

    I've tried to simplify a formula, and I'm not sure if I've done it right. The following is my solution - I've indicated which logic rule(s) I've used to arrive at each line (excuse the rubbish handwriting):

    http://www.zaraphrax.net/~fagg/boolean_mhf.jpg

    Any guidance is appreciated.

    Thanks!
    I think that your solution is ok. You may notice that the last line is in fact $\displaystyle p\Rightarrow q$.

    You can also use truth table to verify that your answer is correct.
    For proposition which have the form of implication there is a shorter way:
    Clearly $\displaystyle p\Rightarrow q$ if false only in one case: p=1, q=0.
    Let us try to the same thing with the first expression
    $\displaystyle [p\land \neg(p\Rightarrow q)] \Rightarrow q$.
    This implication is false only if $\displaystyle [p\land \neg(p\Rightarrow q)]=1$ and q=0. For a conjunction to be true, you need that both of the operands are true. This means that p=1, and $\displaystyle \neg(p\Rightarrow q)$ is also true for p=1,q=0.
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  3. #3
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    Quote Originally Posted by kompik View Post
    I think that your solution is ok. You may notice that the last line is in fact $\displaystyle p\Rightarrow q$.

    You can also use truth table to verify that your answer is correct.
    For proposition which have the form of implication there is a shorter way:
    Clearly $\displaystyle p\Rightarrow q$ if false only in one case: p=1, q=0.
    Let us try to the same thing with the first expression
    $\displaystyle [p\land \neg(p\Rightarrow q)] \Rightarrow q$.
    This implication is false only if [tex][p\land \neg(p\Rightarrow q)]=1[/math and q=0.
    Kompik,

    Thanks for your reply. Much appreciated.
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