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Math Help - Boolean Reduction Question

  1. #1
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    Boolean Reduction Question

    Hi everyone

    Question:
    (AB)'(B(B+C))'

    My working:

    =A'B'(B'B'+B'C')
    =(A'B')(B+B'C')
    =(A'B')(B'C')
    =B'(A'+C')

    did i get it right....can someone help me please???

    tq & rgds
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  2. #2
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    Quote Originally Posted by anderson View Post
    Hi everyone

    Question:
    (AB)'(B(B+C))'

    My working:

    =A'B'(B'B'+B'C')
    =(A'B')(B+B'C')
    =(A'B')(B'C')
    =B'(A'+C')

    did i get it right....can someone help me please???

    tq & rgds
    No, I think that your converting (B(B+C))' to B+B'C' is wrong already. Instead, I get the following

    (B(B+C))'=B'

    This is because we have that B\leq B+C and because from U\leq V we can always conclude that UV=U (and that U+V=V).
    In our case, this means that from B\leq B+C we may conclude that B(B+C)=B.

    Similarly I believe that your converting (AB)' to A'B' is wrong. By De Morgan's Laws you get, instead, that (AB)'=A'+B'.

    (Actually, I feel slightly uneasy about your notation. I'm more the "logical chap" comming from the propositional calculus.)

    If I'm not mistaken, the given expression is equivalent to B' in your notation (or \overline{B} in my notation) - but do the rest of the calculation yourself...
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  3. #3
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    pls help me...

    HI everyone

    Question:

    (A.B)'.(B(B+C)' =

    (AB)' = A'+B',B(B+C)=(BC)'
    A' + B' + (BC)'
    =A'+B'+B'+C'
    =A'+C'

    Pls help advise.

    Tq & rgds
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  4. #4
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    This looks alot like digital technology rather than calculus-logic. Tbf, they are the same thing, just that they use a slightly different form of notation. Anyhow! I'm going to give this a try, but I wouldn't bet my life on the solution:

    First, I'd separate it into 2 parts, effectively:
    (A * B)'
    and
    (B (B + C))'

    (A * B)' = (A' + B')

    (B (B + C))' = (B' + (B' * C' ) = (This can be reduced to B' per de Morgan laws, they all have fancier names but I can never remember them)

    So! You get:
    ((A' + B') B') = A'B' + B'B' = A'B' + B' = B' (de Morgan again)

    Did I get it right?
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  5. #5
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    HI 0113

    Your answer looks very accurate,thank you. Tq for all the help & support..any other ideas,please share...

    Tq & rgds
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  6. #6
    Super Member Failure's Avatar
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    Quote Originally Posted by O113 View Post
    So! You get:
    ((A' + B') B') = A'B' + B'B' = A'B' + B' = B' (de Morgan again)

    Did I get it right?
    Well, yes, but you could have got that result more easily, in a single step, like this: (A' + B') B'=B'

    Why? - Let me explain this by converting the whole thing to set theory: (\overline{A}\cup\overline{B})\cap \overline{B}=\overline{B}
    Which is to say: if you intersect the superset \overline{A}\cup\overline{B} with its subset \overline{B} you get ... just the subset. Or, to use new symbols: from Y\subseteq X you can conclude that X\cap Y= Y (and that X\cup Y=X). In other words, \cap behaves like a greatest lower-bound and \cup behaves like a least upper-bound.

    Or to explain it in propositional calculus: (\neg A\vee \neg B)\wedge \neg B\Leftrightarrow \neg B, because this follows from \neg A\vee \neg B\Rightarrow \neg B. Again: \wedge behaves like a greatest lower-bound and \vee behaves like a least upper-bound.

    I don't know about the symbolism the OP uses. Maybe one should write that (A'+B')B'=B' holds, because B'\leq A'+B'? - But I just don't know how the relationship \subseteq for sets and \Rightarrow for propositions is treated in his formalism.
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  7. #7
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    * = AND
    A ring with a cross in it (can't find the latex command) = Exclusive Or
    + = OR
    ' = Inverse

    The rest you kinda skip since they aren't needed for basic circuit technology. You rewrite logical expressions by using a = instead of caring about =>, ->, <=>
    Example:

    (A + B) means A or B
    A * B means A and B
    A' means "Not A"
    The rest you can figure out for yourself xD

    I can't say that I really know enough of this to actually give you a good explanation about how the arrows are used for more difficult problems. I've always assumed that it's the same as in calculus (loosely based on a short notation in an introductory course)
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