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Math Help - logic question

  1. #1
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    logic question

    I have to show "a" occurs if & only if, whenever "b" occurs, "c" occurs.

    For the only if direction I'm pretty sure this means if I have to assume "a" & "b" and then use this to prove "c".

    But for the if direction, is it correct that I have to assume "c" & "b" and use this to prove "a"?

    Thanks in advance
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  2. #2
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    the if part can be written (b -> c) -> a.
    Since you can rewrite (b -> c) as (!b | c)
    to prove the if part you have to prove
    !b -> a
    c -> a
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  3. #3
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    Thanks, can you define your "!" and "|" symbols for me?
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  4. #4
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by RanDom View Post
    I have to show "a" occurs if & only if, whenever "b" occurs, "c" occurs.

    For the only if direction I'm pretty sure this means if I have to assume "a" & "b" and then use this to prove "c".

    But for the if direction, is it correct that I have to assume "c" & "b" and use this to prove "a"?

    Thanks in advance
    (b\,\Rightarrow\,c) is equivalent to (\neg\,b\ \mbox{or}\ c). So to prove (b\,\Rightarrow\,c)\,\implies\,a, you have to show that both \neg\,b\,\implies\,a and c\,\implies\,a.
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  5. #5
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    i used a notation similar to the one of the java programming language
    & = logical conjunction (∧)
    ! = logical negation ()
    | = logical disjunction (∨)
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  6. #6
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    A little off topic, - how would you show that, (for the only if direction), a -> (b ->c) is equivalent to (a & b) -> c ?

    And just to clarify,
    Quote Originally Posted by RanDom
    for the if direction ... I have to assume "c" & "b" and use this to prove "a"
    is incorrect, right?
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  7. #7
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    Quote Originally Posted by TheAbstractionist View Post
    (b\,\Rightarrow\,c) is equivalent to (\neg\,b\ \mbox{or}\ c). So to prove (b\,\Rightarrow\,c)\,\implies\,a, you have to show that both \neg\,b\,\implies\,a and c\,\implies\,a.
    Wait... I know for a fact that c implies a is false... !!!
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  8. #8
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    always consider x -> y is equivalent to !x | y

    so (a & b) -> c
    is equivalent to !(a & b) | c
    and this is equivalent to !a | !b | c

    also a -> (b ->c)
    is equivalent to !a | (!b | c)
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  9. #9
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by RanDom View Post
    Wait... I know for a fact that c implies a is false... !!!
    It can’t be false if (b\,\Rightarrow\,c)\,\implies\,a is to hold.

    The implication is equivalent to \neg\,(b\,\Rightarrow\,c) or a.

    \neg\,(b\,\Rightarrow\,c) is equivalent to \neg\,(\neg\,b\ \mbox{or}\ c) which is equivalent to b\ \mbox{and}\ \neg\,c.

    Hence the implication is equivalent to (b\ \mbox{and}\ \neg\,c) or a. It follows that if c is true, then (b\ \mbox{and}\ \neg\,c) is not true and so a must be true; in other words, c\,\implies\,a.
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  10. #10
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    Yep nevermind... there was a subtle difference for c->a and the result I knew was false... Thanks for your help guys
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  11. #11
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    Hello, RanDom!

    Have interpreted the question correctly?


    I have to show "a" occurs if & only if, whenever "b" occurs, "c" occurs.
    I could be wrong . . .


    It says: . b \to c)" alt="a \:\longleftrightarrow \b \to c)" />


    So we must prove: . \begin{array}{c}a \longrightarrow (b \to c) \\ \text{and} \\ (b \to c) \longrightarrow a \end{array}

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