boolean-algebrapropositional-calculus
Can this Boolean expression:
$$A*\overline{A*B}$$
be expanded to give:
$$A*\overline{A} * A*\overline{B}$$
Although that appears to reduce to zero?
I know $A(\overline{A+B})$ can be expanded to give: $A*A + A*\overline{B}$
So can it work with an AND? How else do you simplify the first expression?
If your expression is $¬(A \oplus B) \oplus (B \vee ¬C)$, then
$¬(A \oplus B)=1\oplus A\oplus B$, $¬C=1\oplus C$ and $B\vee\neg C=B\oplus(\neg C)\oplus B(\neg C)$ so
$¬(A \oplus B) \oplus (B \vee ¬C)=1\oplus A\oplus B\oplus B\oplus 1\oplus C\oplus B(1+C)=$ $A\oplus B\oplus C\oplus BC=$
$=A\oplus(B\vee C)$
Since $\oplus$ (XOR) is associative, and distributive over $\cdot$ (AND), and $X\oplus X=0$.
($0$ is FALSE and $1$ is TRUE)
The elements $\displaystyle\bigoplus_{i}A^{a_i}B^{b_i}C^{c_i}$, where $a_1,b_i,c_i=0,1$, is the elements of a Boolean ring and is very straightforward to manipulate the algebraic way.
Best Answer
No. By De Morgan's laws, $$(A * B)' = A' + B'.$$ So, $A*(A*B)'$ can be expanded to give $$A*(A'+B') = A*A' + A*B' = \mathsf{F} + A*B' = A*B'.$$