I need help in simplifying a Boolean expression.












0












$begingroup$


My starting point was (A+D)*(A+B+C)*(~A+C+~D)



And I should end at ~A*B*D +A*~D +C*D (according to online solvers.)



But when I do it by hand on a paper I end up with this:
~A*B*D +A*~D +C*D +A*C



And for the life of me I can't figure out how to simplify and get rid of the extra +A*C






boolean-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Does $p+q$ mean $pland q$? Does $pq$ mean $plor q$?
    $endgroup$
    – J.G.
    Dec 9 '18 at 16:37










  • $begingroup$
    + is OR, * is AND, ~is NOT; p+q was p∨q and pq was p∧q if i recall correctly
    $endgroup$
    – Andres Eelma
    Dec 9 '18 at 16:45












  • $begingroup$
    Thanks for clarifying.
    $endgroup$
    – J.G.
    Dec 9 '18 at 16:48










  • $begingroup$
    My bad, I didn't realize using asteriks to show multiplication would make the text italics and whatnot.
    $endgroup$
    – Andres Eelma
    Dec 9 '18 at 16:54










  • $begingroup$
    use pland q for $pland q$, and use plor q for $plor q$, and use sim for $sim$ :)
    $endgroup$
    – Nosrati
    Dec 9 '18 at 16:55


















0












$begingroup$


My starting point was (A+D)*(A+B+C)*(~A+C+~D)



And I should end at ~A*B*D +A*~D +C*D (according to online solvers.)



But when I do it by hand on a paper I end up with this:
~A*B*D +A*~D +C*D +A*C



And for the life of me I can't figure out how to simplify and get rid of the extra +A*C






boolean-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Does $p+q$ mean $pland q$? Does $pq$ mean $plor q$?
    $endgroup$
    – J.G.
    Dec 9 '18 at 16:37










  • $begingroup$
    + is OR, * is AND, ~is NOT; p+q was p∨q and pq was p∧q if i recall correctly
    $endgroup$
    – Andres Eelma
    Dec 9 '18 at 16:45












  • $begingroup$
    Thanks for clarifying.
    $endgroup$
    – J.G.
    Dec 9 '18 at 16:48










  • $begingroup$
    My bad, I didn't realize using asteriks to show multiplication would make the text italics and whatnot.
    $endgroup$
    – Andres Eelma
    Dec 9 '18 at 16:54










  • $begingroup$
    use pland q for $pland q$, and use plor q for $plor q$, and use sim for $sim$ :)
    $endgroup$
    – Nosrati
    Dec 9 '18 at 16:55
















0












0








0





$begingroup$


My starting point was (A+D)*(A+B+C)*(~A+C+~D)



And I should end at ~A*B*D +A*~D +C*D (according to online solvers.)



But when I do it by hand on a paper I end up with this:
~A*B*D +A*~D +C*D +A*C



And for the life of me I can't figure out how to simplify and get rid of the extra +A*C






boolean-algebra







share|cite|improve this question











$endgroup$




My starting point was (A+D)*(A+B+C)*(~A+C+~D)



And I should end at ~A*B*D +A*~D +C*D (according to online solvers.)



But when I do it by hand on a paper I end up with this:
~A*B*D +A*~D +C*D +A*C



And for the life of me I can't figure out how to simplify and get rid of the extra +A*C






boolean-algebra




boolean-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 16:58









Rócherz

2,7762721




2,7762721










asked Dec 9 '18 at 16:35









Andres EelmaAndres Eelma

33




33








  • 1




    $begingroup$
    Does $p+q$ mean $pland q$? Does $pq$ mean $plor q$?
    $endgroup$
    – J.G.
    Dec 9 '18 at 16:37










  • $begingroup$
    + is OR, * is AND, ~is NOT; p+q was p∨q and pq was p∧q if i recall correctly
    $endgroup$
    – Andres Eelma
    Dec 9 '18 at 16:45












  • $begingroup$
    Thanks for clarifying.
    $endgroup$
    – J.G.
    Dec 9 '18 at 16:48










  • $begingroup$
    My bad, I didn't realize using asteriks to show multiplication would make the text italics and whatnot.
    $endgroup$
    – Andres Eelma
    Dec 9 '18 at 16:54










  • $begingroup$
    use pland q for $pland q$, and use plor q for $plor q$, and use sim for $sim$ :)
    $endgroup$
    – Nosrati
    Dec 9 '18 at 16:55
















  • 1




    $begingroup$
    Does $p+q$ mean $pland q$? Does $pq$ mean $plor q$?
    $endgroup$
    – J.G.
    Dec 9 '18 at 16:37










  • $begingroup$
    + is OR, * is AND, ~is NOT; p+q was p∨q and pq was p∧q if i recall correctly
    $endgroup$
    – Andres Eelma
    Dec 9 '18 at 16:45












  • $begingroup$
    Thanks for clarifying.
    $endgroup$
    – J.G.
    Dec 9 '18 at 16:48










  • $begingroup$
    My bad, I didn't realize using asteriks to show multiplication would make the text italics and whatnot.
    $endgroup$
    – Andres Eelma
    Dec 9 '18 at 16:54










  • $begingroup$
    use pland q for $pland q$, and use plor q for $plor q$, and use sim for $sim$ :)
    $endgroup$
    – Nosrati
    Dec 9 '18 at 16:55










1




1




$begingroup$
Does $p+q$ mean $pland q$? Does $pq$ mean $plor q$?
$endgroup$
– J.G.
Dec 9 '18 at 16:37




$begingroup$
Does $p+q$ mean $pland q$? Does $pq$ mean $plor q$?
$endgroup$
– J.G.
Dec 9 '18 at 16:37












$begingroup$
+ is OR, * is AND, ~is NOT; p+q was p∨q and pq was p∧q if i recall correctly
$endgroup$
– Andres Eelma
Dec 9 '18 at 16:45






$begingroup$
+ is OR, * is AND, ~is NOT; p+q was p∨q and pq was p∧q if i recall correctly
$endgroup$
– Andres Eelma
Dec 9 '18 at 16:45














$begingroup$
Thanks for clarifying.
$endgroup$
– J.G.
Dec 9 '18 at 16:48




$begingroup$
Thanks for clarifying.
$endgroup$
– J.G.
Dec 9 '18 at 16:48












$begingroup$
My bad, I didn't realize using asteriks to show multiplication would make the text italics and whatnot.
$endgroup$
– Andres Eelma
Dec 9 '18 at 16:54




$begingroup$
My bad, I didn't realize using asteriks to show multiplication would make the text italics and whatnot.
$endgroup$
– Andres Eelma
Dec 9 '18 at 16:54












$begingroup$
use pland q for $pland q$, and use plor q for $plor q$, and use sim for $sim$ :)
$endgroup$
– Nosrati
Dec 9 '18 at 16:55






$begingroup$
use pland q for $pland q$, and use plor q for $plor q$, and use sim for $sim$ :)
$endgroup$
– Nosrati
Dec 9 '18 at 16:55












1 Answer
1






active

oldest

votes


















0












$begingroup$

By applying the distributive law you get that $(A lor D) land (A lor B lor C)land ( sim A lor C , lor sim D)$ is equivalent to



$$
(A land A land sim A) lor (A land A land C) lor (A land A land sim D) lor (A land B land sim A) lor (A land B land C) lor (A land B land sim D) lor (A land C land sim A) lor (A land C land C) lor (A land C land sim D) lor (D land A land sim A) lor (D land A land C) lor (D land A land sim D) lor (D land B land sim A) lor (D land B land C) lor (D land B land sim D) lor (D land C land sim A) lor (D land C land C) lor (D land C land sim D)
$$



We now delete all the disjuncts that are equivalent to $0$ and we get rid of all the redundant terms. We get that the original expression is equivalent to



$$
(A land C) lor (A land sim D) lor (A land B land C) lor (A land B land sim D) lor (A land C ) lor (A land C land sim D) lor (D land A land C) lor (D land B land sim A) lor (D land B land C) lor (D land C land sim A) lor (D land C)
$$



We now erase the disjuncts that are smaller than another disjunct.



$$
(A land C) lor (A land sim D) lor (D land B land sim A) lor (D land C)
$$



Which, by commutativity of $land$, is equivalent to the expression you got. Therefore your solution was right.



Note that your solution is equivalent to $ (sim A land B land D) lor (A land sim D) lor (C land D)$. Indeed, by the distributive law and the fact that $D , lor sim D$ is equivalent to $1$, your solution is equivalent to
$$
(sim A land B land D) lor (A land sim D) lor (C land D) lor (A land C land D) lor (A land C land sim D).
$$



Since the two last disjuncts are smaller than the third and second ones, it is equivalent to



$$
(sim A land B land D) lor (A land sim D) lor (C land D) .
$$






share|cite|improve this answer









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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

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    active

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    active

    oldest

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    0












    $begingroup$

    By applying the distributive law you get that $(A lor D) land (A lor B lor C)land ( sim A lor C , lor sim D)$ is equivalent to



    $$
    (A land A land sim A) lor (A land A land C) lor (A land A land sim D) lor (A land B land sim A) lor (A land B land C) lor (A land B land sim D) lor (A land C land sim A) lor (A land C land C) lor (A land C land sim D) lor (D land A land sim A) lor (D land A land C) lor (D land A land sim D) lor (D land B land sim A) lor (D land B land C) lor (D land B land sim D) lor (D land C land sim A) lor (D land C land C) lor (D land C land sim D)
    $$



    We now delete all the disjuncts that are equivalent to $0$ and we get rid of all the redundant terms. We get that the original expression is equivalent to



    $$
    (A land C) lor (A land sim D) lor (A land B land C) lor (A land B land sim D) lor (A land C ) lor (A land C land sim D) lor (D land A land C) lor (D land B land sim A) lor (D land B land C) lor (D land C land sim A) lor (D land C)
    $$



    We now erase the disjuncts that are smaller than another disjunct.



    $$
    (A land C) lor (A land sim D) lor (D land B land sim A) lor (D land C)
    $$



    Which, by commutativity of $land$, is equivalent to the expression you got. Therefore your solution was right.



    Note that your solution is equivalent to $ (sim A land B land D) lor (A land sim D) lor (C land D)$. Indeed, by the distributive law and the fact that $D , lor sim D$ is equivalent to $1$, your solution is equivalent to
    $$
    (sim A land B land D) lor (A land sim D) lor (C land D) lor (A land C land D) lor (A land C land sim D).
    $$



    Since the two last disjuncts are smaller than the third and second ones, it is equivalent to



    $$
    (sim A land B land D) lor (A land sim D) lor (C land D) .
    $$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      By applying the distributive law you get that $(A lor D) land (A lor B lor C)land ( sim A lor C , lor sim D)$ is equivalent to



      $$
      (A land A land sim A) lor (A land A land C) lor (A land A land sim D) lor (A land B land sim A) lor (A land B land C) lor (A land B land sim D) lor (A land C land sim A) lor (A land C land C) lor (A land C land sim D) lor (D land A land sim A) lor (D land A land C) lor (D land A land sim D) lor (D land B land sim A) lor (D land B land C) lor (D land B land sim D) lor (D land C land sim A) lor (D land C land C) lor (D land C land sim D)
      $$



      We now delete all the disjuncts that are equivalent to $0$ and we get rid of all the redundant terms. We get that the original expression is equivalent to



      $$
      (A land C) lor (A land sim D) lor (A land B land C) lor (A land B land sim D) lor (A land C ) lor (A land C land sim D) lor (D land A land C) lor (D land B land sim A) lor (D land B land C) lor (D land C land sim A) lor (D land C)
      $$



      We now erase the disjuncts that are smaller than another disjunct.



      $$
      (A land C) lor (A land sim D) lor (D land B land sim A) lor (D land C)
      $$



      Which, by commutativity of $land$, is equivalent to the expression you got. Therefore your solution was right.



      Note that your solution is equivalent to $ (sim A land B land D) lor (A land sim D) lor (C land D)$. Indeed, by the distributive law and the fact that $D , lor sim D$ is equivalent to $1$, your solution is equivalent to
      $$
      (sim A land B land D) lor (A land sim D) lor (C land D) lor (A land C land D) lor (A land C land sim D).
      $$



      Since the two last disjuncts are smaller than the third and second ones, it is equivalent to



      $$
      (sim A land B land D) lor (A land sim D) lor (C land D) .
      $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        By applying the distributive law you get that $(A lor D) land (A lor B lor C)land ( sim A lor C , lor sim D)$ is equivalent to



        $$
        (A land A land sim A) lor (A land A land C) lor (A land A land sim D) lor (A land B land sim A) lor (A land B land C) lor (A land B land sim D) lor (A land C land sim A) lor (A land C land C) lor (A land C land sim D) lor (D land A land sim A) lor (D land A land C) lor (D land A land sim D) lor (D land B land sim A) lor (D land B land C) lor (D land B land sim D) lor (D land C land sim A) lor (D land C land C) lor (D land C land sim D)
        $$



        We now delete all the disjuncts that are equivalent to $0$ and we get rid of all the redundant terms. We get that the original expression is equivalent to



        $$
        (A land C) lor (A land sim D) lor (A land B land C) lor (A land B land sim D) lor (A land C ) lor (A land C land sim D) lor (D land A land C) lor (D land B land sim A) lor (D land B land C) lor (D land C land sim A) lor (D land C)
        $$



        We now erase the disjuncts that are smaller than another disjunct.



        $$
        (A land C) lor (A land sim D) lor (D land B land sim A) lor (D land C)
        $$



        Which, by commutativity of $land$, is equivalent to the expression you got. Therefore your solution was right.



        Note that your solution is equivalent to $ (sim A land B land D) lor (A land sim D) lor (C land D)$. Indeed, by the distributive law and the fact that $D , lor sim D$ is equivalent to $1$, your solution is equivalent to
        $$
        (sim A land B land D) lor (A land sim D) lor (C land D) lor (A land C land D) lor (A land C land sim D).
        $$



        Since the two last disjuncts are smaller than the third and second ones, it is equivalent to



        $$
        (sim A land B land D) lor (A land sim D) lor (C land D) .
        $$






        share|cite|improve this answer









        $endgroup$



        By applying the distributive law you get that $(A lor D) land (A lor B lor C)land ( sim A lor C , lor sim D)$ is equivalent to



        $$
        (A land A land sim A) lor (A land A land C) lor (A land A land sim D) lor (A land B land sim A) lor (A land B land C) lor (A land B land sim D) lor (A land C land sim A) lor (A land C land C) lor (A land C land sim D) lor (D land A land sim A) lor (D land A land C) lor (D land A land sim D) lor (D land B land sim A) lor (D land B land C) lor (D land B land sim D) lor (D land C land sim A) lor (D land C land C) lor (D land C land sim D)
        $$



        We now delete all the disjuncts that are equivalent to $0$ and we get rid of all the redundant terms. We get that the original expression is equivalent to



        $$
        (A land C) lor (A land sim D) lor (A land B land C) lor (A land B land sim D) lor (A land C ) lor (A land C land sim D) lor (D land A land C) lor (D land B land sim A) lor (D land B land C) lor (D land C land sim A) lor (D land C)
        $$



        We now erase the disjuncts that are smaller than another disjunct.



        $$
        (A land C) lor (A land sim D) lor (D land B land sim A) lor (D land C)
        $$



        Which, by commutativity of $land$, is equivalent to the expression you got. Therefore your solution was right.



        Note that your solution is equivalent to $ (sim A land B land D) lor (A land sim D) lor (C land D)$. Indeed, by the distributive law and the fact that $D , lor sim D$ is equivalent to $1$, your solution is equivalent to
        $$
        (sim A land B land D) lor (A land sim D) lor (C land D) lor (A land C land D) lor (A land C land sim D).
        $$



        Since the two last disjuncts are smaller than the third and second ones, it is equivalent to



        $$
        (sim A land B land D) lor (A land sim D) lor (C land D) .
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 8:14









        Luca CaraiLuca Carai

        31119




        31119






























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