How to Simplify Boolean Algebra: Can Someone Help Me with This Problem?

[naspek]

hey there..
can someone help me with the problem given..

Problem..
simplify b ⊕ c'

Homework Equations

The Attempt at a Solution

i don't know how to attempt this question..
can someone can give me an idea to start..
thanks in advance

 

[Mark44]

hey there..
can someone help me with the problem given..

Problem..
simplify b ⊕ c'

What was the symbol between b and c'? Kind of hard to help if we don't know what the problem is.

Homework Equations

The Attempt at a Solution

i don't know how to attempt this question..
can someone can give me an idea to start..
thanks in advance

 

[Zryn]

It looks like an exclusive OR to me, as in 'one or the other but not both':

A B X
0 0 0
0 1 1
1 0 1
1 1 0

So A XOR B = A'B + AB'

So what does that tell you about B XOR C'?

 

[Mark44]

Zryn,
How do you get XOR out of ⊕?

 

[Zryn]

That was the symbol I was introduced to, and used, throughout my University life, essentially a + inside a circle.

Doing a Google search for "XOR Symbol" confirmed this.

If we assume that the symbol is part of a Rorschach test, I'm going with 'XOR Symbol', but what do you see =P?

 

[Mark44]

Maybe it's my browser - I don't see anything but an empty box.

 

[Zryn]

It's a pretty shabby XOR symbol, as far as XOR symbol's go really, but it does the job.

 

[Mark44]

Yeah, all I see is an empty box.

 

[naspek]

It looks like an exclusive OR to me, as in 'one or the other but not both':

A B X
0 0 0
0 1 1
1 0 1
1 1 0

So A XOR B = A'B + AB'

So what does that tell you about B XOR C'?

thank u for the such an idea..
so..
when A XOR B' = AB + A'B'

correct?

 

[Zryn]

Yes, you can algebraically replace each B with B', so A'B + AB' becomes A'B' + A(B')' = A'B' + AB.

Proof below just for completeness:

A B (B') X
0 0 (1) 1
0 1 (0) 0
1 0 (1) 0
1 1 (0) 1

So A XOR B' = A'B' + AB.

All good!

 

[naspek]

Yes, you can algebraically replace each B with B', so A'B + AB' becomes A'B' + A(B')' = A'B' + AB.

Proof below just for completeness:

A B (B') X
0 0 (1) 1
0 1 (0) 0
1 0 (1) 0
1 1 (0) 1

So A XOR B' = A'B' + AB.

All good!

thanks Zrynn =)