Solving this equation to be 3 literals












0












$begingroup$


I am trying to solve this to be 3 literals, but I keep getting errors:



Question:$(x'y' + z)' + z + xy + wz$



My answer:



$(x+y)z' + xy + z(1+w)implies xz' + yz' + xy + zimplies zx + zz' + yz' + xy$



$implies zx + yz' + xy implies xyz + xzz' + xy implies xyz + xy implies yx (1 + z)implies yx$



The right answer should be $x + y + z$






boolean-algebra







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$endgroup$












  • $begingroup$
    I don't understand what you did to get from the second line to the third line ..
    $endgroup$
    – Matti P.
    Dec 20 '18 at 7:26










  • $begingroup$
    I distributed the z to the xz'
    $endgroup$
    – Maysa Sarsour
    Dec 20 '18 at 7:30






  • 2




    $begingroup$
    So that should be $(z+x)(z+z')$, instead of $zx+zz'$.
    $endgroup$
    – Poypoyan
    Dec 20 '18 at 7:32










  • $begingroup$
    Use the @Poypoyan method to simplify the rest of the equation to achieve the expected answer.
    $endgroup$
    – Hoseyn Heydari
    Dec 20 '18 at 7:57
















0












$begingroup$


I am trying to solve this to be 3 literals, but I keep getting errors:



Question:$(x'y' + z)' + z + xy + wz$



My answer:



$(x+y)z' + xy + z(1+w)implies xz' + yz' + xy + zimplies zx + zz' + yz' + xy$



$implies zx + yz' + xy implies xyz + xzz' + xy implies xyz + xy implies yx (1 + z)implies yx$



The right answer should be $x + y + z$






boolean-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't understand what you did to get from the second line to the third line ..
    $endgroup$
    – Matti P.
    Dec 20 '18 at 7:26










  • $begingroup$
    I distributed the z to the xz'
    $endgroup$
    – Maysa Sarsour
    Dec 20 '18 at 7:30






  • 2




    $begingroup$
    So that should be $(z+x)(z+z')$, instead of $zx+zz'$.
    $endgroup$
    – Poypoyan
    Dec 20 '18 at 7:32










  • $begingroup$
    Use the @Poypoyan method to simplify the rest of the equation to achieve the expected answer.
    $endgroup$
    – Hoseyn Heydari
    Dec 20 '18 at 7:57














0












0








0


1



$begingroup$


I am trying to solve this to be 3 literals, but I keep getting errors:



Question:$(x'y' + z)' + z + xy + wz$



My answer:



$(x+y)z' + xy + z(1+w)implies xz' + yz' + xy + zimplies zx + zz' + yz' + xy$



$implies zx + yz' + xy implies xyz + xzz' + xy implies xyz + xy implies yx (1 + z)implies yx$



The right answer should be $x + y + z$






boolean-algebra







share|cite|improve this question











$endgroup$




I am trying to solve this to be 3 literals, but I keep getting errors:



Question:$(x'y' + z)' + z + xy + wz$



My answer:



$(x+y)z' + xy + z(1+w)implies xz' + yz' + xy + zimplies zx + zz' + yz' + xy$



$implies zx + yz' + xy implies xyz + xzz' + xy implies xyz + xy implies yx (1 + z)implies yx$



The right answer should be $x + y + z$






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 20 '18 at 7:39









Yadati Kiran

1,7951619




1,7951619










asked Dec 20 '18 at 7:18









Maysa SarsourMaysa Sarsour

1




1












  • $begingroup$
    I don't understand what you did to get from the second line to the third line ..
    $endgroup$
    – Matti P.
    Dec 20 '18 at 7:26










  • $begingroup$
    I distributed the z to the xz'
    $endgroup$
    – Maysa Sarsour
    Dec 20 '18 at 7:30






  • 2




    $begingroup$
    So that should be $(z+x)(z+z')$, instead of $zx+zz'$.
    $endgroup$
    – Poypoyan
    Dec 20 '18 at 7:32










  • $begingroup$
    Use the @Poypoyan method to simplify the rest of the equation to achieve the expected answer.
    $endgroup$
    – Hoseyn Heydari
    Dec 20 '18 at 7:57


















  • $begingroup$
    I don't understand what you did to get from the second line to the third line ..
    $endgroup$
    – Matti P.
    Dec 20 '18 at 7:26










  • $begingroup$
    I distributed the z to the xz'
    $endgroup$
    – Maysa Sarsour
    Dec 20 '18 at 7:30






  • 2




    $begingroup$
    So that should be $(z+x)(z+z')$, instead of $zx+zz'$.
    $endgroup$
    – Poypoyan
    Dec 20 '18 at 7:32










  • $begingroup$
    Use the @Poypoyan method to simplify the rest of the equation to achieve the expected answer.
    $endgroup$
    – Hoseyn Heydari
    Dec 20 '18 at 7:57
















$begingroup$
I don't understand what you did to get from the second line to the third line ..
$endgroup$
– Matti P.
Dec 20 '18 at 7:26




$begingroup$
I don't understand what you did to get from the second line to the third line ..
$endgroup$
– Matti P.
Dec 20 '18 at 7:26












$begingroup$
I distributed the z to the xz'
$endgroup$
– Maysa Sarsour
Dec 20 '18 at 7:30




$begingroup$
I distributed the z to the xz'
$endgroup$
– Maysa Sarsour
Dec 20 '18 at 7:30




2




2




$begingroup$
So that should be $(z+x)(z+z')$, instead of $zx+zz'$.
$endgroup$
– Poypoyan
Dec 20 '18 at 7:32




$begingroup$
So that should be $(z+x)(z+z')$, instead of $zx+zz'$.
$endgroup$
– Poypoyan
Dec 20 '18 at 7:32












$begingroup$
Use the @Poypoyan method to simplify the rest of the equation to achieve the expected answer.
$endgroup$
– Hoseyn Heydari
Dec 20 '18 at 7:57




$begingroup$
Use the @Poypoyan method to simplify the rest of the equation to achieve the expected answer.
$endgroup$
– Hoseyn Heydari
Dec 20 '18 at 7:57










1 Answer
1






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oldest

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1












$begingroup$

It is often quite useful to consider the negation of the given expression and then use




  • $aa = a$

  • $aa'=0$

  • $a+ab = a(1+b) = a$


So, consider $E' = left( (x'y' + z)' + z + xy + wz right)'$:
$$begin{eqnarray*} E'
& = & left( (x'y' + z)' + z + xy + wz right)' \
& = & (x'y' + z)z'(xy)'(wz)' \
& stackrel{zz' = 0}{=}& x'y'z'(x'+y')(w'+z') \
& stackrel{x'x' = x' etc.}{=}& x'y'z' + x'y'z'w' \
& = & x'y'z'(1+ w' ) \
& = & x'y'z' \
end{eqnarray*}
$$
$$Rightarrow E = (x'y'z')' = x+y+z$$






share|cite|improve this answer









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    1 Answer
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    1












    $begingroup$

    It is often quite useful to consider the negation of the given expression and then use




    • $aa = a$

    • $aa'=0$

    • $a+ab = a(1+b) = a$


    So, consider $E' = left( (x'y' + z)' + z + xy + wz right)'$:
    $$begin{eqnarray*} E'
    & = & left( (x'y' + z)' + z + xy + wz right)' \
    & = & (x'y' + z)z'(xy)'(wz)' \
    & stackrel{zz' = 0}{=}& x'y'z'(x'+y')(w'+z') \
    & stackrel{x'x' = x' etc.}{=}& x'y'z' + x'y'z'w' \
    & = & x'y'z'(1+ w' ) \
    & = & x'y'z' \
    end{eqnarray*}
    $$
    $$Rightarrow E = (x'y'z')' = x+y+z$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      It is often quite useful to consider the negation of the given expression and then use




      • $aa = a$

      • $aa'=0$

      • $a+ab = a(1+b) = a$


      So, consider $E' = left( (x'y' + z)' + z + xy + wz right)'$:
      $$begin{eqnarray*} E'
      & = & left( (x'y' + z)' + z + xy + wz right)' \
      & = & (x'y' + z)z'(xy)'(wz)' \
      & stackrel{zz' = 0}{=}& x'y'z'(x'+y')(w'+z') \
      & stackrel{x'x' = x' etc.}{=}& x'y'z' + x'y'z'w' \
      & = & x'y'z'(1+ w' ) \
      & = & x'y'z' \
      end{eqnarray*}
      $$
      $$Rightarrow E = (x'y'z')' = x+y+z$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        It is often quite useful to consider the negation of the given expression and then use




        • $aa = a$

        • $aa'=0$

        • $a+ab = a(1+b) = a$


        So, consider $E' = left( (x'y' + z)' + z + xy + wz right)'$:
        $$begin{eqnarray*} E'
        & = & left( (x'y' + z)' + z + xy + wz right)' \
        & = & (x'y' + z)z'(xy)'(wz)' \
        & stackrel{zz' = 0}{=}& x'y'z'(x'+y')(w'+z') \
        & stackrel{x'x' = x' etc.}{=}& x'y'z' + x'y'z'w' \
        & = & x'y'z'(1+ w' ) \
        & = & x'y'z' \
        end{eqnarray*}
        $$
        $$Rightarrow E = (x'y'z')' = x+y+z$$






        share|cite|improve this answer









        $endgroup$



        It is often quite useful to consider the negation of the given expression and then use




        • $aa = a$

        • $aa'=0$

        • $a+ab = a(1+b) = a$


        So, consider $E' = left( (x'y' + z)' + z + xy + wz right)'$:
        $$begin{eqnarray*} E'
        & = & left( (x'y' + z)' + z + xy + wz right)' \
        & = & (x'y' + z)z'(xy)'(wz)' \
        & stackrel{zz' = 0}{=}& x'y'z'(x'+y')(w'+z') \
        & stackrel{x'x' = x' etc.}{=}& x'y'z' + x'y'z'w' \
        & = & x'y'z'(1+ w' ) \
        & = & x'y'z' \
        end{eqnarray*}
        $$
        $$Rightarrow E = (x'y'z')' = x+y+z$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 20 '18 at 8:02









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