Using Redundancy to Simplify Boolean Expression

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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...

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2 Answers

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Here is one reduction:

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

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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)

$z+x'y$

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