Boolean Logic Simplification: AB+A'B'=?

Question

I have some problem simplifying the equation AB+A'B'= ? Is there any straight forward answer like AB'+A'B = A^B ; (^)=XOR sign ?

Solution

It's the XNOR function, which typically doesn't occur often enough to warrant its own operator. It's the negation of XOR, though, which can be seen by applying De Morgan's law twice.

AB + A'B' = ((AB)'(A'B')')'
          = ((A' + B')(A + B))'
          = (A'A + B'A + A'B + B'B)'
          = (0 + AB' + A'B + 0)'
          = (AB' + A'B)'
          = (A ^ B)'

or by simply comparing truth tables

A  B    A ^ B   AB + A'B'
-------------------------
0  0      0        1
0  1      1        0
1  0      1        0
1  1      0        1

(Put another way, XNOR is equivalence for two arguments, so you can think of A XNOR B as an operator that converts comparison to a value. A XNOR B equals 1 if A == B is true, 0 if it is false.)

Comments

[Tennyson H]

This is not a programming question, rather it is a Math question. Please ask it on math.stackexchange.com with the logic tag.

[Ridowan Ahmed]

yeah, seems you're right . I'm a CSE student and I've found this problem on my CSE231 course named Digital Logic. I'm was eager to find the solution so I din't give it much thought. I will be careful next time .

[Stranger Forever]

I think there is a minor mistake in this answer. AA'=0 and not 1.

[chepner]

Indeed; not sure what I was thinking, but the following step was done correctly, as if I had typed 0 instead of 1.

[Stranger Forever]

Yes yes, the following steps were all correctly done..