The Boolean Algebra XOR Problem: Are These Expressions Equivalent?

In summary: Which gives you## a (\overline {ab \oplus ac }) + \overline a (ab \oplus ac}) ##You can then use the distributive law to combine like terms:## a (\overline {ab \oplus ac }) + \overline a (ab \oplus ac}) = a \oplus (ab \oplus ac)##which gives you## a (ab \oplus ac) + \overline a (ab \oplus ac}) = a \oplus ab \oplus ac##
  • #1
Rectifier
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The problem
This is not a complete homework problem. I am at the last step of the solution to a long problem and only interested to know whether these following expressions are equivalent.

My answer:
## a \oplus ab \oplus ac ##

Answer in my book:
## a \oplus b \oplus c ##

The attempt
I have tried rewriting that expression and get
## a \oplus ab \oplus ac = a \oplus a(b \oplus c) ##

I am not advancing any longer from there. I am surely missing some algebraic law for boolean expressions. Could you please help me?
 
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  • #2
Did you try building a truth table?
If you do you'll see that your answer is different from the book's answer.
For example ABC = 0,1,0
The book answer gives 1
Your answer results in 0
 
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  • #3
Part of your problem may be your understanding of operator precedence.

It seems to vary slightly across programming languages. For C, Java and Javascript it is:

NOT > AND > XOR > OR

a XOR ( a AND b) XOR ( a AND c)

with XOR equivalent to:

x XOR y = (x OR y) AND (NOT ( x AND y) // mentor note: fixed the expression (thx cpscdave)

reworking the second and third terms should get you to the goal.
 
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  • #4
Well, I guess I am doing something wrong then. :/ Thanks for the help.
 
  • #5
jedishrfu said:
x XOR y = (x OR y) AND (NOT ( x OR y)

Think you mean
(x OR y) AND [NOT ( x AND y)]

or alternatively

(x OR y) AND ( NOT x or NOT y)
 
  • #6
Yes you are right my apologies.

I got my up and down arrows confused in my conversion to ands and ors.

I fixed my post.
 
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  • #7
a ⊕ b ⊕ c does have a rather interesting truth table.
 
  • #8
Rectifier said:
My answer:
## a \oplus ab \oplus ac ##

Answer in my book:
## a \oplus b \oplus c ##

The attempt
I have tried rewriting that expression and get
## a \oplus ab \oplus ac = a \oplus a(b \oplus c) ##

I am not advancing any longer from there. I am surely missing some algebraic law for boolean expressions. Could you please help me?
I would proceed by converting one XOR at a time to its equivalent in AND and OR, starting with this:

## \mathrm {a \oplus ab \oplus ac = a (\overline {ab \oplus ac }) + \overline a (ab \oplus ac)}##
 

1. What is the Boolean Algebra XOR problem?

The Boolean Algebra XOR problem is a logical operation that stands for "exclusive or." It compares two expressions and returns a true value only if one expression is true and the other is false. Otherwise, it returns a false value.

2. Why is the Boolean Algebra XOR problem important?

The Boolean Algebra XOR problem is important because it is a fundamental concept in Boolean algebra and has numerous applications in computer science, mathematics, and engineering. It is used in digital logic circuits, error detection and correction, cryptography, and more.

3. What does it mean for two expressions to be equivalent in Boolean Algebra XOR?

In Boolean Algebra XOR, two expressions are considered equivalent if they produce the same output for all possible input combinations. This means that the expressions have the same logical function and can be substituted for each other without changing the overall result.

4. How do you solve the Boolean Algebra XOR problem?

To solve the Boolean Algebra XOR problem, you can use truth tables, logical equivalences, or algebraic manipulation. A truth table shows all possible input combinations and the resulting output. Logical equivalences are rules that allow you to simplify and transform expressions. Algebraic manipulation involves using Boolean laws and identities to manipulate the expressions until they are equivalent.

5. Can the Boolean Algebra XOR problem be extended to more than two expressions?

Yes, the Boolean Algebra XOR problem can be extended to any number of expressions. For example, the XOR operation between three expressions returns a true value only if an odd number of the expressions are true. Similarly, the XOR operation between four or more expressions follows the same logic.

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