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I was simplifying the equation of a logic gates problem and I realized that ab + bc + cā and ab + cā followed the same truth table which is the following:

-----------------
| A | B | C | S |
-----------------
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 |
-----------------

The problem is that I haven't found any way to get to the simplified expression from the main one by applying boolean algebra procedures. Any idea? Thanks.

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2 Answers 2

up vote 1 down vote accepted

Another way to illustrate this equivalence is a Karnaugh-Veitch map:

enter image description here

The "green" term $b c$ can be eliminated, because it is covered by the disjunction of $c a'$ and $a b$.

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Thanks Alex! Very clear explanation. I didn't thought about justifying it through Karnaught's map. –  IOS_DEV May 25 '13 at 17:33

Note that $b \land c$ is the disjoint union of $b \land c \land a$ and $b \land c \land \bar a$. The first of these is contained in $a \land b$, while the second is contained in $c \land \bar a$.

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