Minimize term without Karnaugh mapsimplifying using Boolean Algebra.Boolean function simplificationThree Boolean Algebra Proofs - I just don't get it!Boolean algebra question.Minimizing using a Karnaugh map when given as subscripts F4,2655How many minimized forms can a boolean expression have with 4 variables?How can i simplyfy this boolean equation?Simplifying a four term Boolean expression using Boolean algebraboolean algebra: simplify 3-term dnf form covering a|~aCircuit to Karnaugh map

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Minimize term without Karnaugh map


simplifying using Boolean Algebra.Boolean function simplificationThree Boolean Algebra Proofs - I just don't get it!Boolean algebra question.Minimizing using a Karnaugh map when given as subscripts F4,2655How many minimized forms can a boolean expression have with 4 variables?How can i simplyfy this boolean equation?Simplifying a four term Boolean expression using Boolean algebraboolean algebra: simplify 3-term dnf form covering a|~aCircuit to Karnaugh map













1












$begingroup$


I have the following term, that should get minimized with Boolean algebra (no Karnaugh map!):



(a ∧ ¬b ∧ c) ∨ (a ∧ c ∧ d) ∨ (b ∧ d)



I already figured out, that the minimzed term is as follows (everything without the middle term) and it makes totally sense to me:



(a ∧ ¬b ∧ c) ∨ (b ∧ d)



However when I try to get same result with boolean algebra I am not able to reproduce it.



My first step is the following:



(a ∧ c) ∧ (¬b ∨ d) ∨ (b ∧ d)



When I apply distributivity-law I end up with the first (non-minimized) term.



Edit: Besides the question above, more general: Can every term that is minimizable by a Karnaugh map be minimized with boolean algebra?










share|cite|improve this question









New contributor




F.M.F. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    What does KV mean?
    $endgroup$
    – Jens
    yesterday










  • $begingroup$
    The german abbreviation for the Karnaugh map... updated the question!
    $endgroup$
    – F.M.F.
    yesterday










  • $begingroup$
    Where is the Boolean algebra? Looks like propositional logic to me.
    $endgroup$
    – Rodrigo de Azevedo
    yesterday















1












$begingroup$


I have the following term, that should get minimized with Boolean algebra (no Karnaugh map!):



(a ∧ ¬b ∧ c) ∨ (a ∧ c ∧ d) ∨ (b ∧ d)



I already figured out, that the minimzed term is as follows (everything without the middle term) and it makes totally sense to me:



(a ∧ ¬b ∧ c) ∨ (b ∧ d)



However when I try to get same result with boolean algebra I am not able to reproduce it.



My first step is the following:



(a ∧ c) ∧ (¬b ∨ d) ∨ (b ∧ d)



When I apply distributivity-law I end up with the first (non-minimized) term.



Edit: Besides the question above, more general: Can every term that is minimizable by a Karnaugh map be minimized with boolean algebra?










share|cite|improve this question









New contributor




F.M.F. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    What does KV mean?
    $endgroup$
    – Jens
    yesterday










  • $begingroup$
    The german abbreviation for the Karnaugh map... updated the question!
    $endgroup$
    – F.M.F.
    yesterday










  • $begingroup$
    Where is the Boolean algebra? Looks like propositional logic to me.
    $endgroup$
    – Rodrigo de Azevedo
    yesterday













1












1








1





$begingroup$


I have the following term, that should get minimized with Boolean algebra (no Karnaugh map!):



(a ∧ ¬b ∧ c) ∨ (a ∧ c ∧ d) ∨ (b ∧ d)



I already figured out, that the minimzed term is as follows (everything without the middle term) and it makes totally sense to me:



(a ∧ ¬b ∧ c) ∨ (b ∧ d)



However when I try to get same result with boolean algebra I am not able to reproduce it.



My first step is the following:



(a ∧ c) ∧ (¬b ∨ d) ∨ (b ∧ d)



When I apply distributivity-law I end up with the first (non-minimized) term.



Edit: Besides the question above, more general: Can every term that is minimizable by a Karnaugh map be minimized with boolean algebra?










share|cite|improve this question









New contributor




F.M.F. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I have the following term, that should get minimized with Boolean algebra (no Karnaugh map!):



(a ∧ ¬b ∧ c) ∨ (a ∧ c ∧ d) ∨ (b ∧ d)



I already figured out, that the minimzed term is as follows (everything without the middle term) and it makes totally sense to me:



(a ∧ ¬b ∧ c) ∨ (b ∧ d)



However when I try to get same result with boolean algebra I am not able to reproduce it.



My first step is the following:



(a ∧ c) ∧ (¬b ∨ d) ∨ (b ∧ d)



When I apply distributivity-law I end up with the first (non-minimized) term.



Edit: Besides the question above, more general: Can every term that is minimizable by a Karnaugh map be minimized with boolean algebra?







propositional-calculus boolean-algebra






share|cite|improve this question









New contributor




F.M.F. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




F.M.F. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited yesterday









Rodrigo de Azevedo

13k41960




13k41960






New contributor




F.M.F. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked yesterday









F.M.F.F.M.F.

1084




1084




New contributor




F.M.F. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





F.M.F. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






F.M.F. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • $begingroup$
    What does KV mean?
    $endgroup$
    – Jens
    yesterday










  • $begingroup$
    The german abbreviation for the Karnaugh map... updated the question!
    $endgroup$
    – F.M.F.
    yesterday










  • $begingroup$
    Where is the Boolean algebra? Looks like propositional logic to me.
    $endgroup$
    – Rodrigo de Azevedo
    yesterday
















  • $begingroup$
    What does KV mean?
    $endgroup$
    – Jens
    yesterday










  • $begingroup$
    The german abbreviation for the Karnaugh map... updated the question!
    $endgroup$
    – F.M.F.
    yesterday










  • $begingroup$
    Where is the Boolean algebra? Looks like propositional logic to me.
    $endgroup$
    – Rodrigo de Azevedo
    yesterday















$begingroup$
What does KV mean?
$endgroup$
– Jens
yesterday




$begingroup$
What does KV mean?
$endgroup$
– Jens
yesterday












$begingroup$
The german abbreviation for the Karnaugh map... updated the question!
$endgroup$
– F.M.F.
yesterday




$begingroup$
The german abbreviation for the Karnaugh map... updated the question!
$endgroup$
– F.M.F.
yesterday












$begingroup$
Where is the Boolean algebra? Looks like propositional logic to me.
$endgroup$
– Rodrigo de Azevedo
yesterday




$begingroup$
Where is the Boolean algebra? Looks like propositional logic to me.
$endgroup$
– Rodrigo de Azevedo
yesterday










1 Answer
1






active

oldest

votes


















2












$begingroup$

beginalign
aoverlinebc+acd+bd & = aoverlinebc + acd(b+overlineb) + bd \
& = aoverlinebc + acdb+acdoverlineb + bd \
& = (aoverlinebc + aoverlinebcd) + (bd + bdac) \
& = aoverlinebc + bd \
endalign






share|cite|improve this answer









$endgroup$












    Your Answer





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    1 Answer
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    oldest

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    2












    $begingroup$

    beginalign
    aoverlinebc+acd+bd & = aoverlinebc + acd(b+overlineb) + bd \
    & = aoverlinebc + acdb+acdoverlineb + bd \
    & = (aoverlinebc + aoverlinebcd) + (bd + bdac) \
    & = aoverlinebc + bd \
    endalign






    share|cite|improve this answer









    $endgroup$

















      2












      $begingroup$

      beginalign
      aoverlinebc+acd+bd & = aoverlinebc + acd(b+overlineb) + bd \
      & = aoverlinebc + acdb+acdoverlineb + bd \
      & = (aoverlinebc + aoverlinebcd) + (bd + bdac) \
      & = aoverlinebc + bd \
      endalign






      share|cite|improve this answer









      $endgroup$















        2












        2








        2





        $begingroup$

        beginalign
        aoverlinebc+acd+bd & = aoverlinebc + acd(b+overlineb) + bd \
        & = aoverlinebc + acdb+acdoverlineb + bd \
        & = (aoverlinebc + aoverlinebcd) + (bd + bdac) \
        & = aoverlinebc + bd \
        endalign






        share|cite|improve this answer









        $endgroup$



        beginalign
        aoverlinebc+acd+bd & = aoverlinebc + acd(b+overlineb) + bd \
        & = aoverlinebc + acdb+acdoverlineb + bd \
        & = (aoverlinebc + aoverlinebcd) + (bd + bdac) \
        & = aoverlinebc + bd \
        endalign







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        JensJens

        3,90021031




        3,90021031




















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