Question

Simplify the following Boolean expression W'X(Z'+YZ)+X(W+Y'Z)

Answer 1

Answer:

After figuring things out with a Karnaugh map, we see that splitting up the xy term will be helpful:

xy+xy′z+x′yz′=xy(1)+xy′z+x′yz′=xy(z+z′)+xy′z+x′yz′=(xyz+xyz′)+xy′z+x′yz′=(xyz+xy′z)+(xyz′+x′yz′)=xz(y+y′)+yz′(x+x′)=xz(1)+yz′(1)=xz+yz′

Explanation:

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Answer 2

WY+YZ′+W′XY′Z would be the final equation.

Simplification:
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Lets simplify the terms,

Terms 1 and 3 give us  W′YZ′ .

Terms 1 and 4 give us  X′YZ′ .

Terms 3 and 6 give us  XYZ′ .

Terms 4 and 5 give us  WX′Y .

Terms 4 and 6 give us  WYZ′ .

Terms 5 and 7 give us  WYZ .

Terms 6 and 7 give us  WXY .

Now we have 8 terms (we need to include the original term 2 which we haven’t used):

W′YZ′+X′YZ′+XYZ′+WX′Y+WYZ′+WYZ+WXY + W′XY′Z

And now we do more work:

Terms 1 and 5 give us  YZ′ .

Terms 2 and 3 give us  YZ′ .

Terms 4 and 7 give us  WY .

Terms 5 and 6 give us  WY .

So now we’re left with

WY+YZ′+W′XY′Z

WY+YZ′+W′XY′Z will be our equation as final result.


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