F = A’B’C’ + ABC’ – (1)
F = A’B + BC’ + A’B – (2)
In function1, each term is called a minterm( A’B’C). The sum of minterms is called a Canonical Sum of Product. It can be directly taken from the Truth Table for the function.
The function2 has terms called a product term which may have one or more literal. The sum of all such terms is called a Standard Sum of Product.
The same concept applies for Canonical Product of Sum form.
Now the problem is that if you are given a Standard Sum of Product for Boolean minimization, what to do, because it appears that its already minimized.
How to simplify the function 2 using K-Map?
Create a three variable map like we do always .and put 1 in the box where a particular term is missing .
Which term is missing ? yes right! it’s C
A’B( C + C’)
A’BC + A’BC’
In the map you put one in both A’BC + A’BC’.
Similarly, Solve other missing variable in each of the terms of the function.
Now , you can minimize the above 3-variable K-Map easily because you have identified the correct boxes.