Hi I'm a student and I got stuck on a problem.

I need to create a logic circuit only using AND gates or NOT gates. But the question in boolean form includes the OR operator and i'm having difficulty figuring out how to get rid of the OR in the expression.

The boolean expression is ABC + A~B~c + AB~C

I know I'll probably have to use the boolean properties to expand/simplify, i've been trying different things but I have no ideas and completely lost.

  • 1
    \$\begingroup\$ Uhh you don't know how to create an or gate from not gates and and gates? \$\endgroup\$ – Bradman175 Feb 4 '17 at 5:52

You just need to know that \$A+B=\overline{\overline{A+B}}=\overline{\overline{A}\cdot\overline{B}}\$. So:

$$\begin{align*} A\cdot B\cdot C + A\cdot\overline{B}\cdot\overline{C} + A\cdot B\cdot\overline{C} &\\\\ &=\overline{\overline{A\cdot B\cdot C + A\cdot\overline{B}\cdot\overline{C} + A\cdot B\cdot\overline{C}}}\\\\ &=\overline{\overline{A\cdot B\cdot C}\cdot\overline{A\cdot\overline{B}\cdot\overline{C}}\cdot\overline{A\cdot B\cdot\overline{C}}} \end{align*}$$

Of course, you could have first simplified your equation. It's just \$A\cdot B + A\cdot\overline{C}\$.

  • \$\begingroup\$ Further simplified to A.(B+'C) \$\endgroup\$ – RoyC Feb 4 '17 at 11:27
  • \$\begingroup\$ @RoyC That was obvious and I'd considered it as a "product of sums" approach. But given the OP I decided to keep the simplified form in "sum of products" form. \$\endgroup\$ – jonk Feb 4 '17 at 16:43
  • \$\begingroup\$ Fair comment but I was going for simplicity \$\endgroup\$ – RoyC Feb 4 '17 at 16:57
  • \$\begingroup\$ @RoyC Yeah. Different goals. No problem. I saw both, though. \$\endgroup\$ – jonk Feb 4 '17 at 17:02

\$ABC + A\overline{B}\overline{C} + AB\overline{C} \$

\$ABC + \color{Red}{AB\overline{C}} + A\overline{B}\overline{C} + AB\overline{C} \$ (Idempotent)

\$AB(C + \overline{C}) + A\overline{C} (\overline{B} + B) \$ (Distributive)

\$AB + A\overline{C}\$ (Complement)

\$\overline{\overline{AB} \cdot \overline{A\overline{C}}}\$ (DeMorgan's)

More NAND than all AND/NOT, but it works.


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