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'
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Sign up to join this communityHow 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'
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')
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.
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.
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
Edit: The answer can be simplified one more step by the boolean algebra law of consensus
If you want to check your work after doing it by hand you could use a program like Logic Friday.
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\$.
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:
Obviously if your input is already in DNF (aka SOP), then obviously the first and second steps don't apply.
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)