SOP expression:
A'B'C'D + A'B'CD' + A'BC'D' + A'BCD + AB'C'D' + AB'CD + ABC'D + ABCD'
Desired expression:
A ⊕ B ⊕ C ⊕ D
Using Boolean algebra, how to convert the SOP expression listed below into A ⊕ B ⊕ C ⊕ D
boolean-algebracomputer sciencelogic
boolean-algebracomputer sciencelogic
SOP expression:
A'B'C'D + A'B'CD' + A'BC'D' + A'BCD + AB'C'D' + AB'CD + ABC'D + ABCD'
Desired expression:
A ⊕ B ⊕ C ⊕ D
Best Answer
Apply the distributive law to factor out $A$ and $A'$:
$$A'(B'C'D + B'CD' + BCD + BC'D') + A(B'C'D' + B'CD + BC'D + BCD')$$
Within the parentheses, factor out $B$ and $B'$:
$$A'(B'(C'D + CD') + B(CD + C'D')) + A(B'(C'D' + CD) + B(C'D + CD'))$$
Rewrite $(C'D+CD')$ by $(C \oplus D)$ and $(C'D'+CD)$ by $(C \oplus D)'$:
$$A'(B'(C \oplus D) + B(C \oplus D)') + A(B'(C \oplus D)' + B(C \oplus D))$$
In a similar manner:
$$A'(B \oplus (C \oplus D)) + A(B \oplus (C \oplus D))'$$
and:
$$A \oplus (B \oplus (C \oplus D))$$
Exclusive or is associative. Therefore, parentheses can be omitted:
$$A \oplus B \oplus C \oplus D$$