Simplify the following Boolean expression. F(W,X, Y, Z) = !WX!Y!Z + WX!Y!Z + !WXY!Z + WXY!Z + !W!XZ + W!X!YZ+W!XYZ
Simplify the following Boolean expression. F(W,X, Y, Z) = !WX!Y!Z + WX!Y!Z + !WXY!Z + WXY!Z + !W!XZ + W!X!YZ+W!XYZ
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![**Simplify the following Boolean expression.**
\[ F(W, X, Y, Z) = \overline{W}X\overline{Y}Z + WX\overline{Y}Z + \overline{W}XY\overline{Z} + WXY\overline{Z} + \overline{W}XZ + W\overline{X}YZ + W\overline{X}YZ \]
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\[ \text{[Input Box for Answer]} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0dbe0800-3ed1-4c93-80b7-6ba53d418546%2F31640877-09db-4a64-9fc9-b99881b69b77%2Fj5cc43d_processed.png&w=3840&q=75)
Transcribed Image Text:**Simplify the following Boolean expression.**
\[ F(W, X, Y, Z) = \overline{W}X\overline{Y}Z + WX\overline{Y}Z + \overline{W}XY\overline{Z} + WXY\overline{Z} + \overline{W}XZ + W\overline{X}YZ + W\overline{X}YZ \]
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\[ \text{[Input Box for Answer]} \]
Expert Solution
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Step 1: To simplify the given Boolean expression F(W, X, Y, Z):
You can use Boolean algebra rules, including simplification using the distributive law and complementation. Here's the simplified expression:
F(W, X, Y, Z) = !WX!Y!Z + WX!Y!Z + !WXY!Z + WXY!Z + !W!XZ + W!X!YZ + W!XYZ
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