Boolean algebra simplification

I need to reduce this boolean expression to its simplest form. Its given that the simplest form contains 3 terms and 7 literals.

The expression is:

x'yz + w'x'z + x'y + wxy + w'y'z

We tried this in class, and even our recitation teacher could not figure it out.

Any help would be appreciated.

4

6 Answers

Try putting it into a Karnaugh Map.

4

Quine-McCluskey reduction is one of the strongest tools for this, although it can be labor-intensive.

I'm a bit rusty with boolean algebra, but I think I've worked out how to do this. I'll let you do the working, but here are the basic steps:

1) Group the terms with y and eliminate what you can inside the brackets. Once expanded again, this will leave you with four terms and ten literals.

2) Eliminate the redundant term, leaving you with three terms and seven literals.

Hint: I first worked out the answer with a Karnaugh map, and then used regular boolean algebra to get to the solution :-)

1

Like this:

x'y + wxy + w'y'z

11

Can we use groups?

w'z(x' + y') + y(x' + w)

1

X'YZ + W'X'Z + X'Y + WXY + W'Y'Z
= X'Y+W'X'Z+WXY+W'Y'Z by absorption
= WY+X'Y+W'X'Z+W'Y'Z by absorption
= W'Y'Z+WY+X'YZ+X'Y by consensus
= W'Y'Z+WY+X'Y by absorption

Using tool at

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