Thank you for posting one question at a time. I appreciate it.
Here are the standard minterms for 2 variables only.
| X | Y | Term | Designation |
| 0 | 0 | X'Y' | m0 |
| 0 | 1 | X'Y | m1 |
| 1 | 0 | XY' | m2 |
| 1 | 1 | XY | m3 |
The given boolean function
F(X,Y) = X + X'Y
needs to be rewritten in such a way that we involve only the minterms listed.
That lone X needs to have a Y or Y' attached to it somehow.
We have to do a bit of trickery to get the function to be a sum of standard minterms.
F(X,Y) = X + X'Y
F(X,Y) = X*1 + X'Y .... identity law
F(X,Y) = X*(Y+Y') + X'Y .... inverse law
F(X,Y) = X*Y + X*Y' + X'Y .... distributive law
At this point, we have the boolean function written as the sum as these specific minterms
m3 = XY
m2 = XY'
m1 = X'Y
If we wanted, we could express that sum as F(X,Y) = m1+m2+m3 = Σ(1,2,3)
This is the blank K-map for the two variables X and Y.
Along the top we have 0 and 1 corresponding to either X' or X respectively.
Along the left side we have 0 and 1 corresponding to either Y' or Y respectively.
These rules apply to minterms only.
Inside the table itself, we'll enter the m0 through m3 as shown below.
Here's some informal notation to help keep track.
m0 = 00 = X'Y'
m1 = 01 = X'Y
m2 = 10 = XY'
m3 = 11 = XY
Or you could use the minterm table mentioned at the top of the page.
Next, we replace m1,m2,m3 with 1's since they are part of the F(X,Y) function after we expanded things out.
The fact that m0 is left out means it gets assigned 0.
Now comes the grouping.
Use these K-Map grouping rules
http://www.ee.surrey.ac.uk/Projects/Labview/minimisation/karrules.html
and you can review on this page if needed
http://www.ee.surrey.ac.uk/Projects/Labview/minimisation/karnaugh.html#introduction
There are a lot of rules to keep in mind, but once you get the hang of it, it's not too bad.
The goal is to group the '1's together in batches of powers of 2. In this case, it means we either want groups of 1 or groups of 2.
The other goal is to minimize the number of groups, while also maximizing the group sizes.
Doing both of those tasks can be achieved with this grouping here.
Groups can overlap.
The blue group represents X because XY'+XY = X(Y'+Y) = X*1 = X
Similarly, the green group represents Y
X'Y+XY = (X'+X)Y = 1*Y = Y
Therefore, we end up with X+Y as the final simplification.
Here's the boolean verification table
| X | Y | X’ | X’Y | X+X’Y | X+Y |
| 0 | 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 | 1 |
The last two columns for X+X'Y and X+Y are the same.
This proves that X+X'Y = X+Y
Answer: X+Y