Simplifying output for a XOR gate using Boolean Algebra

  • #1

Homework Statement


I'm trying to show that the output of this XOR circuit is ##F=A'B+AB'##,
480px-XOR_from_NOR.svg.png


Homework Equations


##(A+B)'=A'\cdot B'##
##(A\cdot B)'=A'+B'##

The Attempt at a Solution



From the gates the output is ##[(A\cdot B)+(A+B)']'##, using De Morgan's laws this becomes ##[(A\cdot B)+(A+B)']'=(A\cdot B)'\cdot (A+B)=(A\cdot B)'\cdot (A+B)=(A'+B')\cdot (A+B)=0##? I cant seem to figure out what I'm doing wrong.
 

Answers and Replies

  • #2
ehild
Homework Helper
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Homework Statement


I'm trying to show that the output of this XOR circuit is ##F=A'B+AB'##,
480px-XOR_from_NOR.svg.png


Homework Equations


##(A+B)'=A'\cdot B'##
##(A\cdot B)'=A'+B'##

The Attempt at a Solution



From the gates the output is ##[(A\cdot B)+(A+B)']'##, using De Morgan's laws this becomes ##[(A\cdot B)+(A+B)']'=(A\cdot B)'\cdot (A+B)=(A\cdot B)'\cdot (A+B)=(A'+B')\cdot (A+B)=0##? I cant seem to figure out what I'm doing wrong.
You did it right, but simplify the last expression applying the distributive law.
 
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  • #3
You did it right, but simplify the last expression applying the distributive law.
yikes, can't believe I missed that one!
 
  • #4
ehild
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1,901
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