Please help with all of them and show work please Simplify the following expressionsF(x, y, z) = xy + x’y’z’ + x’yz’F(x, y, z) = x’yz + xy’z + x’yz’F(x, y, z) = xy’z’ + xz + x’y’zF(w, x, y, z) =x’z + w’xy’ + w(x’y + xy’)F(w, x, y, z) =w’x’y’z’ + wy’z’ + x’yz’ + w’xyz + xy’z

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Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Please help with all of them and show work please

Simplify the following expressions

1. F(x, y, z) = xy + x’y’z’ + x’yz’
2. F(x, y, z) = x’yz + xy’z + x’yz’
3. F(x, y, z) = xy’z’ + xz + x’y’z
4. F(w, x, y, z) =x’z + w’xy’ + w(x’y + xy’)
5. F(w, x, y, z) =w’x’y’z’ + wy’z’ + x’yz’ + w’xyz + xy’z

Expert Solution

Step 1

F(x, y, z) = xy + x’y’z’ + x’yz’

Using Boolean algebra, we can simplify the expression using De Morgan's laws and the distributive property.

xy + x’y’z’ + x’yz’

= xy + x’(y’z’) + (x’y)z’ (distributive property)

= xy + (x’ + x’y)z’ (De Morgan's laws)

= xy + x’z’ (De Morgan's laws)

So, the simplified expression is: F(x, y, z) = xy + x’z’

F(x, y, z) = x’yz + xy’z + x’yz’

Using Boolean algebra, we can simplify the expression using the distributive property.

x’yz + xy’z + x’yz’

= (x’y + xy’)z + x’yz’ (distributive property)

= z(x’y + xy’) + x’yz’ (distributive property)

= z + x’yz’ (De Morgan's laws)

So, the simplified expression is: F(x, y, z) = z + x’yz’

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