Boolean Expression Evaluation: A XOR (B XOR C) = A XOR (BC' + B'C)

[spaghetti3451]

Homework Statement

Evaluate the following: A XOR (B XOR C) = A XOR (BC' + B'C). Using Boolean expressions.

[This question came up in my test today (as part of a review of 'Digital Logic Design') for my 'Computer Architecture' course. It was worth 6 marks. The total test was marked out of 15.]

Homework Equations

Laws of Boolean Algebra

The Attempt at a Solution

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B XOR C = BC' + B'C, by definition. So, I don't see what there is to show anyway. :(

 

[NascentOxygen]

It's not clear just what is intended, but perhaps you are being asked to remove the XOR operators from the LHS expression. The RHS can then be regarded as a hint to help you on your way.

Your answer will have no XOR operation.

I'd mark the examiner's question, " Must do better on next test... 1/5"

 

[spaghetti3451]

Thank you so much for your kind opinion.

Thus far at my university, I have never once complained to myself about any rude verbal comments (from my lecturers) or difficult test papers (set by some lecturers). It's better to adapt than to complain. That's been my personal motto. And it's served me well. I have consistently got A grades in all my courses.

But, I could not find it in my heart and soul not to complain about this question set in my test today. :(

To make matters worse, my lecturer said I would not have to read textbooks for the test. His notes will do. His notes only had a 3-variable binary counter, and he asked us to implement a 4-variable decimal counter in the exam out of thin air. I guess he was expecting a miracle of genius from his students, i.e. that they would somehow manages to realize that AND gates are needed in the circuit to limit the count to 10 and then reset to 0.

 

[spaghetti3451]

Say the question was:

Show/Prove the following: A XOR (B XOR C) = A XOR (BC' + B'C). Using Boolean expressions.

I still don't see if it could be solved, though.

 

[NascentOxygen]

Say the question was:

Show/Prove the following: A XOR (B XOR C) = A XOR (BC' + B'C). Using Boolean expressions.

I still don't see if it could be solved, though.

I don't like the question. Period. But I guess you could construct a Truth Table, hence demonstrating that RHS = LHS, all the while pretending that you didn't recognize BC' + B'C as being equivalent to XOR.

EDIT: ignore what I just wrote, I overlooked the 'Using Boolean expressions' directive. Though in this context, I don't know what it means, actually.