How to simplify Boolean expression with XOR?

Question

I want to simplify a boolean Expression.
The Expression is something like this

X1 xor (X2 || X3 &&  X4 || x5)

How do I simplify this expression using rules of Boolean Algebra.

Moreover I want to convert the above boolean expression to a CNF form , so how do I do it.

Solution 1

If you consider that A XOR B is equivalent to (A AND ~B) OR (~A AND B) then the rest should be easy.

Solution 2

Write a xor b as:

a != b

or:

!(!a || b) && !(a || !b)

Comments

[Admin]

:-I don't get u what will be the simplification in the above case bcoz in my case A is X1 and B is (X2 || X3 && X4 || x5) so how will this be done can u please explain in detail with a example.

[Draken]

Your Ais X1 and your B is (X2 || X3 && X4 || x5), can you try and work it out from that? Just plug it into the formula @PaulR has provided, simplify and you're good to go

[Paul R]

@Draken: I sincerely hope that the OP is not still struggling with this 4 years after he asked the question! ;-)

[Draken]

True @PaulR, I was editing an answer from someone else and hadn't looked at the date of the question. Thought it was a new one, ah well :)

[Paul R]

@Draken: yes, I'm not sure why someone decided to add another answer at this late stage - probably just trying to build up some rep.