How to convert an expression from SOP to POS and back in Boolean Algebra?

Question

How to convert a Sum of Products (SOP) expression to Product of Sums (POS) form and vice versa in Boolean Algebra?

e.g.: F = xy' + yz'

Answer 1

I think the easiest way is to convert to a k-map, and then get the POS. In your example, you've got:

  \ xy
 z \  00    01    11    10
    +-----+-----+-----+-----+
 0  |     |  x  |  x  |  x  |
    +-----+-----+-----+-----+
 1  |     |     |     |  x  |
    +-----+-----+-----+-----+

In this case, excluding the left column gives (x+y), and excluding the two bottom middle boxes gives (z' + y'), giving an answer of (x + y)(z' + y')

Answer 2

F= xy' + yz' it is in SOP form

This can also be soved using Simple Boolean Algebra techniques as:

Applying Distributive Law :- F=( xy') + y . z'

F= (xy' + y).( xy' + z') which is now converted to POS form.

Answer 3

Another method is just take the compliment of the given expression:

As: xy' + yz'

Taking its compliment:
(xy' + yz')'

=(xy')'.(yz')' {Using De Morgans Law's (a+b)'=a'.b'}

=(x'+y)(y'+z)

Which is the POS form of the complement.

Answer 4

Use DeMorgan's law twice.

Apply the law once:

F' = (xy' + yz')'
   = (xy')'(yz')'
   = (x'+y)(y'+z)
   = x'y' + x'z + yy' + yz
   = x'y' + x'z + yz

Apply again:

F=F''
 =(x'y'+x'z+yz)'
 =(x'y')'(x'z)'(yz)'
 =(x+y)(x+z')(y'+z')
 =(x+y)(y'+z')

Verify the answer using wolframalpha.com

xy' + yz'

(x+y)(y'+z')

Edit: The answer can be simplified one more step by the boolean algebra law of consensus

Answer 5

If you want to check your work after doing it by hand you could use a program like Logic Friday.

Answer 6

It is in a minimum/Sum of Products [SOP] and maximum/Product of Sums [POS] terms, so we can use a Karnaugh map (K map) for it.

For SOP, we pair 1 and write the equation of pairing in SOP while that can be converted into POS by pairing 0 in it and writing the equation in POS form.

For example, for SOP if we write \$x \cdot y \cdot z\$ then for pos we write \$x+y+z\$.

Answer 7

See the procedure at Conjunctive Normal Form: Converting from first-order logic.

This procedure covers the more general case of first order logic, but propositional logic is a subset of first order logic.

Simplifying by ignoring first order logic, it's:

• Eliminate implications
• Move negations inwards by applying DeMorgan's law
• Distribute disjunctions over conjunctions

Obviously if your input is already in DNF (aka SOP), then obviously the first and second steps don't apply.

Answer 8

Let x = ab'c + bc'

x' = (ab'c + bc')'

By DeMorgan's theorem, x' = (a' + b + c')(b' + c)

x' = a'b' + a'c + bb' + bc + c'b' + c'c

x' = a'b' + a'c + bc + c'b'

Employing DeMorgan's theorem again, x = (a'b' + a'c + bc + c'b')'

x = (a + b)(a + c')(b' + c')(c + b)