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I came across a question which I need to solve using Redundancy:

x'z + x'y + xy'z + yz

Could anyone help me solving this? I must solve it using Redundancy only...

Donny Avaris
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  • You must be able to use other principles than redundancy ... with redundancy alone you can't simplify this. – Bram28 Oct 23 '16 at 15:33
  • Then how to solve it using other principles? I'm stuck – Donny Avaris Oct 23 '16 at 15:34
  • you can take xy'z + yz and make that into (xy' + y)z (distribution), and then you can simplify xy' + y to x + y (reduction .... different instructors and different texts may use different names though). So you get (x + y)z which you can distribute back to xy + xz, and now you can combine those with x'y and x'z respectively using further laws. What laws are available for you to use? – Bram28 Oct 23 '16 at 15:41
  • The questions didn't limit what laws I'm allowed to use so I guess everything is okay – Donny Avaris Oct 23 '16 at 15:42
  • Good. Interestingly, I just simplified it along the lines I suggested and never used Redundancy ... anyway, good luck, and if you still can't figure it out ask again. – Bram28 Oct 23 '16 at 15:46
  • Hmm I don't quiet understand how to continue combining xy+xz with x'y and x'z. – Donny Avaris Oct 23 '16 at 15:47
  • Let us continue this discussion in chat. – Bram28 Oct 23 '16 at 15:51

2 Answers2

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Here is one reduction:

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

Brian M. Scott
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Here is what I did:

$x'z+x'y+xy'z+yz=$ (Distribution)

$x'z+x'y+(xy'+y)z=$ (Reduction)

$x'z+x'y+(x+y)z=$ (Distribution)

$x'z+x'y+xz+yz=$ (Adjacency)

$z+x'y+yz=$ (Redundancy)

$z+x'y$

Bram28
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