Formal Boolean Proof of A ⊕ B' ⊕ C = (A ⊕ B ⊕ C)

[nahanksh]

Homework Statement

Prove that
$$A \oplus B' \oplus C = (A \oplus B \oplus C)'$$

Homework Equations

The Attempt at a Solution

I tried to use $$A \oplus B' \oplus C$$ = ABC' + A'B'C' + A'BC + AB'C

But i am not sure how to proceed further from there...

Please could someone give me a little bit of help ?

 

[Synaesthesia]

I would start with the right hand side - it can be rewritten with some laws.

 

[Synaesthesia]

I'm sorry that was still vague. De Morgan's laws to be specific.
NOT (P OR Q) = (NOT P) AND (NOT Q)
NOT (P AND Q) = (NOT P) OR (NOT Q)

 

[Synaesthesia]

I'm sorry that was still vague. De Morgan's laws to be specific.
NOT (P OR Q) = (NOT P) AND (NOT Q)
NOT (P AND Q) = (NOT P) OR (NOT Q)

 

[DeeNos]

To prove this statement, we will use the formal rules of Boolean algebra. First, we will use the property that A ⊕ B = (A + B)(A' + B') to expand the left side of the equation:

A ⊕ B' ⊕ C = (A + B')(A' + C) + (A' + B)(B' + C)

Next, we will distribute the terms using the distributive property of Boolean algebra:

= A'A' + A'C + B'A' + B'C + A'B' + A'C + AB' + BC

= A'C + B'A' + B'C + A'B' + AB' + BC

Now, we can use the property that A' + A = 1 to simplify the terms:

= A'C + B'A' + B'C + A'B' + AB' + BC + A'A + B'B + C'C

= A'C + B'A' + B'C + A'B' + AB' + BC + 1

Next, we can use the property that A + A' = 1 to simplify the terms further:

= A'C + B'C + A'B' + BC + 1

Finally, we can use the property that A + A'B = A + B to simplify the terms one last time:

= A'C + B'C + BC + 1

= (A + B)(C + 1)

= (A + B)(1)

= A + B

= (A + B)' (since A + B = A ⊕ B)

Therefore, we have proven that A ⊕ B' ⊕ C = (A ⊕ B ⊕ C)' using the formal rules of Boolean algebra.