How to simplify the following expression :
A'BCD + AB'CD' + AB'CD + ABC'D + ABCD' + ABCD ?
It should get AC + BCD + ABD using Kmap but using boolean algebra i am stuck no matter how i try .
[Math] Simplifying boolean function using boolean algebra
boolean-algebra
Related Solutions
It’s not too hard to check the results after you know them:
$$\begin{align*}B'D+ABC'D&=(B'+ABC')D\\ &=(B'+AB'C'+ABC')D\\ &=\Big(B'+A(B'+B)C')\Big)D\\ &=(B'+AC')D\\ &=B'D+AC'D \end{align*}$$
and
$$\begin{align*}A'BC + B'C + AC&=B'C+(A'B+A)C\\ &=B'C+(A'B+A+AB)C\\ &=B'C+\Big(A+(A'+A)B\Big)C\\ &=B'C+(A+B)C\\ &=B'C+AC+BC\;. \end{align*}$$
That second one can be further simplified to $AC+C=C$, since $B'C+BC=(B'+B)C=C$.
In both calculations I used the absorption law: $B'=B'+AB'C'$ in the first, and $A=A+AB$ in the second.
When you have no more than three or four proposition letters, you may find Venn diagrams helpful.
HINT
This equivalence principle will be your friend:
Adjacency
$PQ + PQ' = P$
If you're not allowed to use Adjacency in $1$ step, here is a derivation of Adjacency in terms of more basic equivalence principles:
$$P Q + (P Q') \overset{Distribution}=$$
$$P (Q + Q') \overset{Complement}=$$
$$P 1 \overset{Identity}=$$
$$P$$
To apply Adjacency, note that $P$ and $Q$ can be any complex expressions, so in this case, where every terms has $4$ variables, just look for two terms that are the same for $3$ of the variables, but differ in the fourth. For example, the first two terms are the same except for the $D$ variable, so those can be combined:
$A'BC'D'+A'BC'D=A'BC'$
You can also combine the first and seventh terms:
$A'BC'D'+ABC'D'=BC'D'$
To do both of those, you would need to 'reuse' the first term, but you can get as many copies as you want by:
Idempotence
$P + P = P$
So, for example, focusing on the first, second, and seventh term:
$$A'BC'D'+A'BC'D+ABC'D'\overset{Idempotence}=$$
$$A'BC'D'+A'BC'D'+A'BC'D+ABC'D'\overset{Commutation}=$$
$$A'BC'D'+A'BC'D+A'BC'D'+ABC'D'\overset{Adjacency \ x \ 2}=$$
$$A'BC'+BC'D'$$
Best Answer
Triplicate ABCD. then reorder:
A'BCD + AB'CD' + AB'CD + ABC'D + ABCD' + ABCD = (A'BCD + ABCD) + (AB'CD' + AB'CD + ABCD' + ABCD) + (ABC'D + ABCD)
Now use A+A'=1 (etc)
A'BCD + ABCD = (A'+A)BCD = BCD
AB'CD' + AB'CD + ABCD' + ABCD = AC (as explained by William)
ABC'D + ABCD = ABD