Here is Prob. 9, Sec. 1.2, in the book Discrete Mathematics and Its Applications by Kenneth H. Rosen, 8th edition:
Are these system specifications consistent? "The system is in multiuser state if and only if it is operating normally. If the system is operating normally, the kernel is functioning. The kernel is not functioning or the system is in interrupt mode. If the system is not in multiuser state, then it is in interrupt mode. The system is not in interrupt mode."
My Attempt:
Let us put
$$
\begin{align}
p \qquad &\colon \mbox{The system is in multiuser state.} \\
q \qquad &\colon \mbox{The system is operating normally.} \\
r \qquad &\colon \mbox{The kernel is functioning.}
\end{align}
$$
Then our specifications are
$$
\begin{align}
1. \qquad & p \leftrightarrow q, \\
2. \qquad & q \rightarrow r, \\
3. \qquad & \overline{r} \lor \overline{q}, \\
4. \qquad & \overline{p} \rightarrow \overline{q}, \\
5. \qquad & \overline{q}.
\end{align}
$$
If $q$ is False, then Spec. 5 becomes True and so do Specs. 4, 3, and 2; finally, when $q$ is False, for the Spec. 1 to become True $p$ must also be False.
Thus all the Specs. 1 through 5 above become True when both $p$ and $q$ are False.
Hence the given system specifications are consistent.
Is this solution correct? If so, is it clear enough and exhaustive? Or, are there issues?
Best Answer
SPECIFICATION :
"The system is in multiuser state if and only if it is operating normally.
If the system is operating normally, the kernel is functioning.
The kernel is not functioning or the system is in interrupt mode.
If the system is not in multiuser state, then it is in interrupt mode.
The system is not in interrupt mode."
$$\begin{align} A \qquad &\colon \mbox{The system is in multiuser state.} \\ B \qquad &\colon \mbox{The system is operating normally.} \\ C \qquad &\colon \mbox{The system is in interrupt mode.} \\ D \qquad &\colon \mbox{The kernel is functioning.} \\ \end{align}$$
$$\begin{align} 1. \qquad & A \leftrightarrow B \\ 2. \qquad & B \rightarrow D \\ 3. \qquad & \overline{D} \lor {C} \\ 4. \qquad & \overline{A} \rightarrow {C} \\ 5. \qquad & \overline{C} \end{align}$$
CONCLUSION :
It is inconsistent.
PROOF / ANALYSIS :
(5) says $\lnot C$ , then (4) gives $\lnot (\lnot A)$ which is $A$
(1) gives $B$
(2) gives $D$
(3) gives $C$
that (3) $C$ is inconsistent with earlier (5) $\lnot C$