If A + ~A = 1 why ~A~B~C~D + ABCD != 1?

Question

In Boolean algebra the law says that $A$ or complement of $A$ is $1$, then why $ABCD$ or its complement $ABCD$ is not also $1$ ?

Because not ABCD is not (not A and not B and not C and not D)! — OmG, Dec 25 '17 at 15:04
Look up the DeMorgan's theorem to understand what not ABCD actually is and why. — Gaurang Tandon, Dec 25 '17 at 15:06

score 1 · Answer 1 · answered Dec 25 '17 at 15:04

We have: $$X + \overline X = 1$$ and for $X=ABCD$, we have: $$(ABCD) + \overline{ABCD}=1$$

But, $$\overline{ABCD} \color{red}{\neq} \overline A \, \overline B \, \overline C \, \overline D$$ Take for example, $A= 0, B = 1, C=0, D= 1$: The LHS is $1$ and the RHS is $0$.

If A + ~A = 1 why ~A*~B*~C*~D + A*B*C*D != 1?

1 Answers1

If A + ~A = 1 why ~A~B~C~D + ABCD != 1?