I hope the title wasn't butchered but I'll do my best to explain here. I am trying to find the simplified function of $F$ using a K-map and after going through everything I arrived at the simplified function of $F=1$ which doesn't seem right so I wanted to run my work through here and see if there is something I'm missing. below is all my steps
$$F=\overline{ABC}+\overline{A}B+AB\overline{C}+AC$$
which results in the K-map:
I hope the K-Map is clear, the three groups below are: $$$$ $$=AB\overline{C}+A\overline{B}C$$ $$=A$$
above is the first one, below is the second,
$$\overline{ABC}+\overline{A}BC$$ $$=\overline{A}$$
and below is the third group,
$$=ABC+AB\overline{C}+\overline{A}BC+\overline{A}B\overline{C}$$ $$=AB+\overline{A}B$$ $$=B$$
After all of these I am left with a simplified function of
$$F=A+\overline{A}+B$$ $$F=1+B$$ since $$A+\overline{A}=1$$ $$F=1$$ since$$ 1+X=1$$
I don't know if this is alright, I hope I was able to get my steps across properly, is it possible to have a simplified function of 1? I know that the original F does not always result in 1, there are cases where F=0, because of that I don't really know what I did wrong.
Could you give an example where $F=0$? As I see it, the function $F$ does reduce to $F=1$ as shown below:
\begin{align} F & =\overline{ABC}+\overline{A}B+AB\overline{C}+AC \\ & = \overline{A}+ \overline{B}+ \overline{C}+\overline{A}B+AB\overline{C}+AC \\ & = (\overline{A}+\overline{A}B)+ \overline{B}+ (\overline{C}+AB\overline{C})+AC \\ & = \overline{A}+ \overline{B}+ \overline{C}+AC \\ &=(\overline{A}+ \overline{C})+\overline{(\overline{A}+ \overline{C})}+ \overline{B} \\ &=1+ \overline{B} \\ &=1 \end{align} The rules used are De Morgan, commutative, absorption, commutative and De Morgan, complementation and finally annihilator.
EDIT
If $\overline{ABC}$ means "Not(A) and Not(B) and Not(C)" (usually written as $\overline{A}\cdot \overline{B}\cdot \overline{C}$) the expression reduces to $$F=B + A\odot C$$ where $\odot$ is the Exclusive NOR.