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I have a question: is this true the way I solve it?

$$\begin{align*} xyz&+xyz'+xy'z+xy'z'+x'y'z+x'yz'\\ &= xy+xy'+x'y'z+x'yz'\\ &= x(y+y')+x'y'z+x'yz'\\ &= x+x'y'z+x'yz'\\ &= x+x'z'z\\ &=x \end{align*}$$

Am I doing something wrong here?

Brian M. Scott
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Donny Avaris
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2 Answers2

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It clearly can’t be right, because the original expression is true when $x$ and $y$ are false and $z$ is true, and the final expression is not.

You were doing fine up through $x+x'y'z+x'yz'$, but the next step doesn’t work. However,

$$\begin{align*} x+x'y'z+x'yz'&=x+x'(y'z+yz')\\ &=\big(x+x(y'z+yz')\big)+x'(y'z+yz')\\ &=x+\big(x(y'z+yz')+x'(y'z+yz')\big)\\ &=x+y'z+yz'\;, \end{align*}$$

and you’re not going to get it any simpler than that.

Brian M. Scott
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  • thx for the correction but I have a question; is it possible to solve the problem with consensus? – Donny Avaris Oct 22 '16 at 13:48
  • @Donny: For the step where you went wrong: no, because in the terms $x'y'z$ and $x'yz'$ both the literal $y$ and the literal $z$ appear in opposition, and in that case you can’t form a consensus term. You can form a consensus term only when there is exactly one literal that appears in opposition. – Brian M. Scott Oct 22 '16 at 13:53
  • I meant using the consensus from the start. Is it possible? Or is it okay to simplify first using absorption then consensus? Sorry, I'm a complete beginner to this and my teacher doesn't teach well... – Donny Avaris Oct 22 '16 at 13:56
  • @Donny: It’s okay to use any legitimate identity at any stage in the computation. – Brian M. Scott Oct 22 '16 at 13:58
  • Thank you very much. But I don't quiet understand about your step of solving you posted before; why there is x(y'z+yz')+x'(y'z+yz')? is it from an identity? – Donny Avaris Oct 22 '16 at 14:00
  • @Donny: I used absorption in reverse: $x=x+x(y'z+yz')$. I just added some parentheses to make that a little clearer. – Brian M. Scott Oct 22 '16 at 14:02
  • Ah i get it now. Thank you for your time – Donny Avaris Oct 22 '16 at 14:03
  • @Donny: You’re welcome. – Brian M. Scott Oct 22 '16 at 14:04
  • Oh and one last thing, could I use consensus for x' + xy' so that it will be y'? – Donny Avaris Oct 22 '16 at 14:06
  • @Donny: No, because $x'+xy'$ is true when $x$ is false and $y$ is true, but $y'$ isn’t. However, $$x'+xy'=(x'+x'y')+xy'=x'+(x'+x)y'=x'+y';.$$ – Brian M. Scott Oct 22 '16 at 14:13
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The following truth-table indicates that you are indeed doing something wrong:

 x | y | z | xyz | xyz' | xy'z | xy'z' | x'y'z | x'yz' | result | equal to x
---|---|---|-----|------|------|-------|-------|-------|--------|------------
 0 | 0 | 0 |  0  |  0   |  0   |  0    |   0   |   0   |   0    |     yes
 0 | 0 | 1 |  0  |  0   |  0   |  0    |   1   |   0   |   1    |     no
 0 | 1 | 0 |  0  |  0   |  0   |  0    |   0   |   1   |   1    |     no
 0 | 1 | 1 |  0  |  0   |  0   |  0    |   0   |   0   |   0    |     yes
 1 | 0 | 0 |  0  |  0   |  0   |  1    |   0   |   0   |   1    |     yes
 1 | 0 | 1 |  0  |  0   |  1   |  0    |   0   |   0   |   1    |     yes
 1 | 1 | 0 |  0  |  1   |  0   |  0    |   0   |   0   |   1    |     yes
 1 | 1 | 1 |  1  |  0   |  0   |  0    |   0   |   0   |   1    |     yes

An alternative solution:

This expression is an OR of $6$ out of $8$ possible ANDs.

So you can take the negation of an OR of the remaining $2$ ANDs.

The remaining $2$ ANDs are $x'yz$ and $x'y'z'$.

The negation of their OR is $(x'yz+x'y'z')'$.

Using de-morgan law, it is equivalent to $(x+y'+z')(x+y+z)$.

barak manos
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