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I have following truth Table :

F(w,x,y,z) = Σ(0,2,5,7,8,10,12,13,14);

    Truth Table
    W   X   Y   Z   F
0   0   0   0   0   1
1   0   0   0   1   0
2   0   0   1   0   1
3   0   0   1   1   0
4   0   1   0   0   0
5   0   1   0   1   1
6   0   1   1   0   0
7   0   1   1   1   1
8   1   0   0   0   1
9   1   0   0   1   0
10  1   0   1   0   1
11  1   0   1   1   0
12  1   1   0   0   1
13  1   1   0   1   1
14  1   1   1   0   1
15  1   1   1   1   0

With the following Karnaugh Map

  W  X
Y
Z

KMap

I'm able to reduce this to the following SOP

 ~x~z + w~z + x~yz + ~wxy

My Professor explicitly told me that this function can be reduced to 2 SOP terms, but he doesn't have the time to demonstrate this as finals start this week.

I want to trust my professor on this, but I do not see how this can be reduced any more than it is.

How to prove that this can be reduced to 2 terms or prove that it can not be reduced to less than 4 ?

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Not sure if typo. There is a slight mistake in your solution. $$F = \overline x\overline z + w\overline z + x\overline yz+\overline wxz $$

Further reduction is not possible. K-map gives the minimal SOP expression.

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There is an online Karnaugh Map solver here. If you enter your data there on the four input variable page, it comes up with 4 sum of product terms and three product of sum terms.

You get the same result from this logic minimization page, which also prints out the Karnaugh Map.

If 15 was also set, then you could have two product terms as long as you had an active low output. Remember that you can combine terms by wrapping around the ends of the map to the opposite edge.

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