How Can You Simplify This Boolean Expression Using Three Gates?

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SUMMARY

The Boolean expression R=(A · ¬B)+(¬A · B) can be simplified using three gates: one NAND, one AND, and one OR. The solution involves applying de Morgan's theorem to transform the expression into the form (¬A + ¬B) · (A + B), which can then be rewritten as ¬(A · B) · (A + B). This method effectively demonstrates the equivalence of the original expression and the simplified form while adhering to the constraints of using only three gates.

PREREQUISITES
  • Understanding of Boolean algebra principles
  • Familiarity with de Morgan's theorem
  • Knowledge of logic gate functions (NAND, AND, OR)
  • Ability to manipulate logical expressions
NEXT STEPS
  • Study Boolean algebra simplification techniques
  • Learn more about de Morgan's theorem applications
  • Explore the functionality and implementation of NAND gates in digital circuits
  • Investigate practical examples of logic gate combinations in circuit design
USEFUL FOR

Students in electrical engineering, computer science, and anyone involved in digital logic design who seeks to understand Boolean expression simplification and logic gate implementation.

conorordan
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Homework Statement



R=(A \cdot \overline{B})+(\overline{A} \cdot B)​

Using Boolean algebra, including the complement relation D\cdot \overline{D}=0, convert R to a form that uses one NAND, one AND and one OR gate (a total of three gates).

Homework Equations



de Morgan's theorem...

\overline{A+B}=\overline{A}\cdot\overline{B}
\overline{A \cdot B}=\overline{A} + \overline{B}

The Attempt at a Solution



By some trial and error I arrived at this;

(\overline{A}+\overline{B})\cdot(A+B)

And then using the second of de Morgan's equations on the first OR I got to this;

\overline{A\cdot B}\cdot(A+B)

Which works as a solution, it has one NAND, one AND and one OR, however, I can't just skip to that bit, it's the working out in between that I'm missing, I'm not sure how to fill in the gaps.

Thanks for any help you can offer!
 
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You can show that the last statement is equivalent to the original one if you work backwards - it is easier in that direction, but the other direction works as well.
 

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