Simplify this function using Boolean algebra laws?
I am stuck on this part of my homework
(x2'x1'x0') + (x2'x1x0) + (x2x1x0')
I know you have to use distributive laws to do this.
Any help is appreciated.
Thank you.
I have the same question. Someone on stack overflow gave me this answer(i don't know if its correct, because i don't know if we can use the ^:
!(a + b + c) || (b * a ^ c)
where ^ is xor.
you can see my post go to site stackoverflow
search-> Boolean algebra simplification to lowest form
To simplify the given function using Boolean algebra laws, you can follow these steps:
Step 1: Apply the distributive law.
(x2'x1'x0') + (x2'x1x0) + (x2x1x0') = x2'(x1'x0' + x1x0) + x2x1x0'
Step 2: Simplify each term individually.
Term 1: x2'(x1'x0' + x1x0)
Using the distributive law again, you can expand the first term:
= x2'x1'x0' + x2'x1x0
Term 2: x2x1x0'
This term is already fully simplified.
Step 3: Combine the simplified terms:
The two simplified terms can now be combined using the OR operator, as they are connected with a plus sign.
= (x2'x1'x0' + x2'x1x0) + x2x1x0'
Step 4: Observe the pattern and simplify further:
Observe that the common factor in both terms of the first sum (the part inside the parentheses) is x2'. We can factor it out:
= x2'(x1'x0' + x1x0 + x1x0') + x2x1x0'
By applying the distributive law again, we can simplify the expression within the outer parentheses:
= x2'(x1'(x0' + x0) + x1(x0 + x0')) + x2x1x0'
Next, simplify the terms within the inner parentheses:
= x2'(x1'(1) + x1(1)) + x2x1x0'
Simplifying further, we get:
= x2'(x1' + x1) + x2x1x0'
Now apply the Boolean algebra law called complementation:
x1' + x1 = 1, since x and x' are complements.
= x2'(1) + x2x1x0'
Finally, apply the identity law (1st and 2nd terms simplify to their respective variables):
= x2' + x2x1x0'
Therefore, the simplified expression is:
x2' + x2x1x0'