Simplifying equations using boolean algebra

[mrlupr]

I need to simplify these boolean expressions.

Homework Statement

Problem A:
a`b`d+a`bd`+bc`d`+bcd`+ab`c`d+b`cd

Problem B:
a`b`cd+a`bd+abcd+a`b`d+acd+abd+ab`c`d

Problem C:
a`b`c`d`+a`b`cd`+a`bd+ab`c`d`+bd+ab`d`

Homework Equations

abc means a AND b AND c
a+b means a OR b
a` means NOT a

The Attempt at a Solution

I have been working on these for over 6 hours and I still have a mess. I look at the rules and try to apply them, but I know my answers are incorrect.

 

[tiny-tim]

Welcome to PF!

Hi mrlupr! Welcome to PF!

Let's start with …

Problem A:
a`b`d+a`bd`+bc`d`+bcd`+ab`c`d+b`cd

The bc`d`+bcd`part is easy …

what is that? …

and how much further can you get?

Hint: try grouping together all the terms with b`d

 

[mrlupr]

here is what I have so far.

bc`d`+bcd`
bbc`+cd`d`
bc`+c`d
c`+bd

how does it look?

 

[tiny-tim]

here is what I have so far.

bc`d`+bcd`
bbc`+cd`d`
bc`+c`d
c`+bd

how does it look?

sorry … I've no idea what you're doing

 

[mrlupr]

Ok how about this

bc`d`+bcd`
bd`(c`+c)
bd`1

am I any closer?

 

[tiny-tim]

Ok how about this

bc`d`+bcd`
bd`(c`+c)
bd`1

am I any closer?

Yes, that's it!

But write it with = signs, and there's no need to write 1 …

bc`d`+bcd`
= bd`(c`+c)
= bd`

ok, now have a go at all the b`d terms ​

 

[mrlupr]

Ok let's go

a`b`d+a`bd` +bd`+ ab`c`d+b`cd`
=a`11+bd`+ab`d
=a`+ab`d+bd`
=b`d+bd`

How does it look?

 

[tiny-tim]

a`b`d+a`bd` +bd`+ ab`c`d+b`cd`
=a`11+bd`+ab`d

again … no idea what you're doing