1

I have a question to solve the following expression and get it in terms of product of sums

(AB' + A'B)C

And I tried taking the compliment of this 

[(AB' + A'B)C]'

[(AB' + A'B)' + C']

[(AB')'.(A'B)' + C']

[(A' + B).(A + B') + C']

[(A' + B + C').(A + B' + C')]

Is this the correct method and is this the answer? I took the compliment because of the fact that Sum of products are 1's and Products of sums are 0's

enter image description here

Padmal
  • 595
  • Since you started by taking the complement your answer is not equivalent to the original expression, – coffeemath Jan 03 '16 at 12:50
  • Yes but the sum of products is 1 when product of sums is 0. So it doesn't need to be equal? – Padmal Jan 03 '16 at 12:53
  • Maybe make a truth table to compare the original with your final answer. I think your process works provided one initially replaces each literal by its complement, by a duality argument, but that's only a vague memory I had. – coffeemath Jan 03 '16 at 13:07

1 Answers1

1

As @coffeemath noted, your answer is actually the complement of the original expression. So this method is not correct.

I think the easiest way is to use the identities $(x + y)' = x'y'$ and $(xy)' = x' + y'$ (known as De Morgan's laws) and $x'' = x$. We have $$AB' + A'B = ((AB')'(A'B)')' = ((A'+B)(A+B'))'=$$ $$=(A'A + A'B' + AB + BB')' = (AB + A'B')' = (AB)'(A'B')' = (A'+B')(A+B).$$ Hence the original expression equals $$ (AB'+A'B)C = (A'+B')(A+B)C.$$

Alternatively, if you're familiar with how to get a SOP or a POS from the truth table, then you can just construct the truth table for the function $f(A, B)$ realized by the expression $AB' + A'B$ and construct a POS directly from it. In our case the truth table looks like this: $$\begin{array}{|c|c|} \hline A & B & f(A, B) \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \hline \end{array}$$

To construct the POS representation of $f(A, B)$ you should look at pairs $(a, b)$ with $f(a, b) = 0$. With every such pair you associate a sum $A^\sigma + B^\tau$, where $A^\sigma = A$ if $a = 0$ and $A^\sigma = A'$ if $a = 1$, and same for $B^\tau$. After that you just construct the product of all these sums. In our case we have two pairs $(0, 0)$ and $(1, 1)$. Corresponding sums are $A + B$ and $A' + B'$. Finally, we have the POS $(A'+B')(A+B)$.

Random Jack
  • 2,906
  • I kinda understand :) But how to put 'C' into the $(A' + B')(A + B)$ as we need the product of sum of all three terms? – Padmal Jan 03 '16 at 22:08
  • @Blogger: To obtain POS from the original expression you transform only the part of it (namely, $AB' + A'B$) and $C$ stands as it is. Final expression has the form $(A'+B')(A+B)C$ which is a POS ($C$ is just a sum consisting of one summand). – Random Jack Jan 03 '16 at 22:10
  • AB'C + A'BC
  • $\Pi$(2,5)
  • (A+B'+C)(A'+B+C)
  • (A'+B+C')(A+B'+C') These are the multiple choice answers for that
    • – Padmal Jan 03 '16 at 22:14
  • The answer is 1. You just apply the distributive law $(x+y)z = xz + yz$ to the original expression: $$(AB'+A'B)C = AB'C + A'BC.$$ But that is not a product of sums, it is sums of product. Are you sure you need to express it in a POS form? Other answers are incorrect, since your expression equals $0$ at 6 points. – Random Jack Jan 03 '16 at 22:20
  • It is a product of sum I need. I've edited my question to put in a screen. – Padmal Jan 03 '16 at 22:30
  • 1
    It is a different task as I understand. You're supposed to find the POS-form for the complement of the original expression, since you're a given $Y'$ and the question is about $Y$. In that case your approach works and the answer you have got is right (4). – Random Jack Jan 03 '16 at 22:37
  • 1
    Yea it was my bad I didn't mention that important part. I thought because we are taking 1's as SoP and 0's as PoS it doesn't matter. But it does it seems. Thank you very much Jack for the knowledge \m/ – Padmal Jan 03 '16 at 22:44