Convert Prod of Sums to Sum of Prod

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The discussion focuses on converting a product of sums expression into a sum of products format. The initial expression is simplified through various logical laws and transformations, ultimately leading to the expression F = (p'q') + r + s'. An alternative method using a Karnaugh map (K-map) is suggested for efficiency, where the complement of F is computed to identify minterms. The K-map approach confirms the same result, demonstrating the equivalence of the two methods. The conversation emphasizes the importance of simplification techniques in logical expressions.
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Question: Convert to Sum of Products

(p+q’+r+s’). (p’+q+r+s’). (p’+q’+r+s’)

= ((p+q’+r+s’)’ + (p’+q+r+s’)’ + (p’+q’+r+s’)’)’

(Demorgans each Sum-Term)

= ((p’.q.r’.s) + (p.q’.r’.s) + (p.q.r’.s))’ (Factor p.r’.s)

= ((p’.q.r’.s) + (p.r’.s).(q’ + q))’ (Inverse Law)

= ((p’.q.r’.s) + (p.r’.s))’ (Factor r’.s)

= ((r’.s).(p’.q + p))’ (Absorbtion Variant)

= ((r’.s). (q + p))’ (Distributive Law)

= (q.r’.s + p.r’.s)’ (Demorgans outer expression)

= ((q.r’.s)’. (p.r’.s)’) (Demorgans inner expression)

= .(q’+r+s’) . (p’+r+s’)

Still not in SOP
 
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There are many ways to solve this.
The BASIC WAY is to simplify the expression. Let's name the Expression F
F = (p+q’+r+s’). (p’+q+r+s’). (p’+q’+r+s’)
F = (p + q')(p'+q)(p'+q') + (r+s') ---By factoring out (r+s')
F = (pp' + pq + p'q' + qq')(p'+q') + (r+s') ---Multiplying (p + q')(p'+q)
(pq+p'q')(p'+q') + (r+s') --- Some terms are canceled pp' = 0, qq' = 0
(pp'q + pqq' + p'q' + p'q') + (r+s') --Multiplying (pq+p'q')(p'+q')
(p'q') + r + s' --- Some terms are canceled pp'q = 0, pqq' = 0 and p'q' + p'q' = p'q'

F = (p'q') + r + s'THE OTHER METHOD
I think the easiest and most efficient way to solve this is to use a K-map especially in really long expressions.
lets name your function F
Computing for complement of F
F' = ((p+q’+r+s’) + (p’+q+r+s’)’ + (p’+q’+r+s’))'= (p'qr's)(pq'r's)(pqr's)
Then plot it on the K-MAP
(SEE BELOW)Then group the remaining ones (1s)
Which will yield:

F = s'+(p'q') + r.

Which is the same as the above expression.
 

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kukumaluboy said:
Question: Convert to Sum of Products

(p+q’+r+s’). (p’+q+r+s’). (p’+q’+r+s’)

= ((p+q’+r+s’)’ + (p’+q+r+s’)’ + (p’+q’+r+s’)’)’

(Demorgans each Sum-Term)

= ((p’.q.r’.s) + (p.q’.r’.s) + (p.q.r’.s))’ (Factor p.r’.s)

= ((p’.q.r’.s) + (p.r’.s).(q’ + q))’ (Inverse Law)

= ((p’.q.r’.s) + (p.r’.s))’ (Factor r’.s)

= ((r’.s).(p’.q + p))’ (Absorbtion Variant)

= ((r’.s). (q + p))’ (Distributive Law)

= (q.r’.s + p.r’.s)’ (Demorgans outer expression)

= ((q.r’.s)’. (p.r’.s)’) (Demorgans inner expression)

= .(q’+r+s’) . (p’+r+s’)

Still not in SOP
In order to get sum of minterms you should multiply what you have gotten so far, then simplify. Also, you could've done that at the very beginning. What electronicsguy suggested is preferred; when you complement the maxterms given in the original function, you get positions of zeros in your K-table, then minimize the ones as to get sum of minterms.
 
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