Disjunctive Normal Form (DNF) A Boolean function $f$ of $n$ variables is said to be in a DNF if it is a disjunction of conjunction in $n$ variables. That is, each conjunction includes all the variables of the formula and there are no duplicates. We can build a DNF by looking at a truth table and matching a unique conjunction to each of the truth outputs.
Let $f=\overline{AB}CD+\bar{A}BCD+ABCD+A\bar{B}C\bar{D}$, then its DNF is
$
f=\overline{AB}CD+\bar{A}BCD+ABCD+A\bar{B}C\bar{D}$
$=(\bar{A}+\bar{B})CD+\bar{A}BCD+ABCD+A\bar{B}C\bar{D}$
$=\bar{A}CD(B+\bar{B})+\bar{B}CD(A+\bar{A})+\bar{A}BCD+ABCD+A\bar{B}C\bar{D}$
$=ABCD+\bar{A}BCD+\bar{A}\bar{B}CD+A\bar{B}CD+A\bar{B}C\bar{D}.$
In view of definition of DNF, the number of truth outputs of $f$ must be equal to the number of terms in DNF of $f$. Therefore, the number of truth outputs in this case must be $5$. But I found in an article that the number of truth outputs is $4$, and in that article the combination of the value $A\overline{B}CD$ is missing from the list of combinations leading to a truth output.
I need a confirmation, whether the number of truth outputs of $f$ is $4$ or $5$?
Edited:
This is online generated truth table for the first expression.
Hence it is verified that the number of truth outputs is $5$, but not $4$.
Best Answer
I had do edit your question to understand what was going on. Right from the start, there is a MathJax problem: the commands '\overline{AB}' and '\overline{A}\overline{B}' produce $\overline{AB}$ and $\overline{A}\overline{B}$, respectively, so the output is the same on the screen. One solution could be to use '\bar{A}\bar{B}' in the latter case, which produces the output ${\bar A}{\bar B}$.
Thus your final line should be $$ABCD+\overline{A}BCD+\bar{A}\bar{B}CD+A\overline{B}CD+A\overline{B}C\overline{D}$$ which is now the right DNF.