# K-Map resulting in a simplified function of 1

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.

• It is not true that $\overline{ABC} + \bar ABC = \bar A$. Rather, we have $$\overline{ABC} + \bar ABC = \bar A(\overline{BC} + BC)$$ which amounts to $A$ and $NXOR(B,C)$ – Omnomnomnom Oct 12 '18 at 2:57
• That is, assuming you've gone with the usual convention and used $\overline{BC}$ to denote (not B) and (not C) rather than not (B and C) – Omnomnomnom Oct 12 '18 at 3:01
• @Omnomnomnom I must admit I've never seen the convention you refer to as the usual convention. In my experience, the usual convention is this one. – Jens Oct 12 '18 at 17:23
• @Omnomnomnom well I know that $wy+w\overline{y}=w$ so in this case $w=\overline{A}$ and $y=\overline{BC}$ and $\overline{y}=BC$ making what I did true unless I misunderstood the rule – AFC Oct 12 '18 at 17:27

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.

• Well I drew out the truth table for this question and in the case where $A=0$, $B=0$, and $C=1$ then we have $\overline{ABC}$ evaluate to 0, $\overline{A}B$ is also 0, $AB\overline{C}$ is also 0, and finally, $AC$ is also 0 making F zero, I'm not a hundred percent sure but this means that the simplified F can't be 1 since F can evaluate to 0 – AFC Oct 12 '18 at 17:31
• When $A=0$, $B=0$ and $C=1$ then $\overline{ABC}=1$. – Jens Oct 12 '18 at 17:38
• Are you sure about that? $\overline{ABC}$ is supposed to be Not(A) and Not(B) and Not(C), it is not NOT(ABC). When I punch in Not(A) and Not(B) and Not(C) into Wolfram alpha it tells me that the case of A=F, B=F, C=T results in F=F. – AFC Oct 12 '18 at 17:42
• Where did you see that notation? How would you then show Not(A and B and C)? – Jens Oct 12 '18 at 17:45
• I just messed up, I realized after that I was confusing everyone with my notation, sorry about that. Given that I have clarified my notation, does it make more sense now why I was asking if the simplified F was correct when I got 1 for it. – AFC Oct 12 '18 at 18:01