How Do You Simplify Complex Boolean Expressions?

[mad_monkey_j]

Homework Statement

1.
Simplify the following problem using boolean algebra:
(p+q'r')(p'q'+r)

2.
Use algebra or karnaugh maps to simplify the following circuit:
(z+xy')+yx+xzy'

Homework Equations

Boolean algebra rules and demorgan's law

The Attempt at a Solution

1. (p+q'r')(p'q'+r)
=pq'p'+pr+q'r'p'q'+q'r'r
=0q'+pr+'q'r'p+0q
=pr+q'r'p'
=0r+q'r'
=00+q'
=q'
Not sure if correct

2.(z+xy')+yx+x(z+y')
=zxy'+yx+xzy'
=zxy'+1x+xz
=x(zy'+z)
=xz
Not sure about this one either

 

[LCKurtz]

Homework Statement

1.
Simplify the following problem using boolean algebra:
(p+q'r')(p'q'+r)

2.
Use algebra or karnaugh maps to simplify the following circuit:
(z+xy')+yx+xzy'

Homework Equations

Boolean algebra rules and demorgan's law

The Attempt at a Solution

1. (p+q'r')(p'q'+r)
=pq'p'+pr+q'r'p'q'+q'r'r
=0q'+pr+'q'r'p+0q
=pr+q'r'p'

OK to there. Then next equation is wrong. Use the identity a + a'b = a + b.

=0r+q'r'
=00+q'
=q'
Not sure if correct

2.(z+xy')+yx+x(z+y')
=zxy'+yx+xzy'

What rule did you use to get that step?

 

[mad_monkey_j]

So i get
pr+q' for the first one
and i used demorgan's law to get to the second step in the second question

 

[LCKurtz]

So i get
pr+q' for the first one
and i used demorgan's law to get to the second step in the second question

Actually, after looking at it a bit more leisurely than I did this morning, I think you should have just stopped at pr + q'r'p' which is correct. I don't think my suggestion was good.

For the second one, use the associative law to group the two center terms first to get started.