Simplify the following Boolean expression. Show your work and list which axiom or theorem you used in each step. Your final equation should be in minimized sum-of- products (SOP) form.
M = (A + B'C)(A'D + AB + AB’C + AC')B
How can this be done?
I have: Distribute = (AB+B'CB)(A'D + AB + AB'C + AC')
Combining = (AB + C)(A'D + AB + AB'C + AC')
What could be next? I can't split the four sets in the second parenthesis as they all have different stuff. Is there any further to go?
What does it mean to put something in Minimized SOP form?
Best Answer
Normally you have to set up a Karnaugh map to work out this kind of problem but in this case $$\begin{align}(A+\bar BC)(\bar AD+AB+A\bar BC+A\bar C)B&=(A+\bar BC)B(\bar AD+AB+A\bar BC+A\bar C)\\ &=(AB+B\bar BC)(\bar AD+AB+A\bar BC+A\bar C)\\ &=(AB+0C)(\bar AD+AB+A\bar BC+A\bar C)\\ &=(AB+0)(\bar AD+AB+A\bar BC+A\bar C)\\ &=AB(\bar AD+AB+A\bar BC+A\bar C)\\ &=A\bar ABD+AABB+AAB\bar BC+AAB\bar C\\ &=0BD+AB+A0C+AB\bar C\\ &=0+AB+0+AB\bar C\\ &=AB+AB\bar C\\ &=AB1+AB\bar C\\ &=AB(1+\bar C)\\ &=AB1\\ &=AB\end{align}$$ Where we have used the commutative and, associative, and distributive laws. Also that facts that $0A=0$, $1A=1$, $0+A=A$, $1+A=1$, $AA=A$. and $A\bar A=0$. You can check by verifying that if $A=B=1$ the expression evaluates to $1$ while if $A=0$ or $B=0$ the expression evaluates to $0$.