How can I determine the sop and pos part from a combined expression

Question

(, , , ) = (′ + ′)( + ′)′ + ′(′ + ′) + ′

like this combined expression where SOP and POS are both available how can I know which are sop & which are pos?

I am trying to find the sop & pos from this combined expression but I can't. We know (′ + ′)( + ′) it's a pos function and it's a ′A + ′ sop function. But I can't understand how to find from combined expressions where both sop pos are available.

Answer 1

To determine expression F as sum-of-products, you could rewrite it:

F(A, B, C, D) = (D' + AB')(AD + C')B' + BD'(A' + C') + A'
              = (ADD' + C'D' + AAB'D + AB'C')B' + (A'BD' + BC'D') + A'
              = (AB' + B'C'D' + AB'D + AB'C) + (A'BD' + BC'D') + A'
              = C'D' + B'D + A'

The result is a sum-of products.

The truth-table:

A  B  C  D  |  F
------------+------
0  0  0  0  |  1
0  0  0  1  |  1
0  0  1  0  |  1
0  0  1  1  |  1
0  1  0  0  |  1
0  1  0  1  |  1
0  1  1  0  |  1
0  1  1  1  |  1
1  0  0  0  |  1
1  0  0  1  |  1
1  0  1  0  |  0
1  0  1  1  |  1
1  1  0  0  |  1
1  1  0  1  |  0
1  1  1  0  |  0
1  1  1  1  |  0
------------+-------

Every row with F=1 corresponds to a minterm, a product of inputs (inverted or non-inverted). So, the truth table gives a sum-of-products form.

To get a product-of-sums form, you can take the four rows for F=0 and invert the inputs (cf. De Morgan's laws):

  A  B  C  D  |
--------------+------
  1  0  1  0  |  0
  1  1  0  1  |  0
  1  1  1  0  |  0
  1  1  1  1  |  0
--------------+------
        
  (A'+B+C'+D)(A'+B'+C+D')(A'+B'+C'+D)(A'+B'+C'+D')

Answer 2

(D'              + AB')(AD + C')B'  + BD'(A' +    C') + A'
 D'ADB' + D'C'B' + AB'ADB' + AB'C'B' + BD'A' + BD'C'  + A'  distributive law
 D'ADB' + D'C'B' + AB'ADB' + AB'C'B'         + BD'C'  + A'  absorptive law
 D'ADB' + D'C'B' + AB'D    + AB'C'           + BD'C'  + A'  idempotent law
 0 A B' + D'C'B' + AB'D    + AB'C'           + BD'C'  + A'  complement law
 0      + D'C'B' + AB'D    + AB'C'           + BD'C'  + A'  annulment law
          D'C'B' + AB'D    + AB'C'           + BD'C'  + A'  identity law
          D'C'B'           +  B'D    +  B'C' + BD'C'  + A'  absorptive law  
          B'C'D' + BC'D'   +  B'D    +  B'C'          + A'  commutative law 
            C'D'(B' + B)   +  B'D    +  B'C'          + A'  distributive law  
            C'D'1          +  B'D    +  B'C'          + A'  complement law
            C'D'           +  B'D    +  B'C'          + A'  identity law
            C'D'           +  B'D                     + A'  kmap reduction
(populating kmap in above order, B'C' is alreay included when we get to it)

kmap
BC\D  0 1
00    1 1
01    0 1
11    0 0
10    1 0