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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?

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    $\begingroup$ It looks like you viewed "in interrupt mode" as the negation of "operating normally". That may be a feature of common language involving real world machines, but it isn't built into the mere logic of the statements. $\endgroup$
    – coffeemath
    Jan 18 at 8:29
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    $\begingroup$ Note that in user Prem's answer, he symbolizes interrupt mode with a fourth letter. Doing that makes the translations straightforward. $\endgroup$
    – coffeemath
    Jan 18 at 8:44
  • $\begingroup$ Correct , the System may well be operating normally , with Multiuser State , while in interrupt mode , @coffeemath $\endgroup$
    – Prem
    Jan 18 at 8:45

1 Answer 1

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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) $\lnot C$

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  • $\begingroup$ If (3) is treated as a regular OR, as it is denoted here, then True OR True = True, but also True OR False = True. So we can have a functioning kernel and also be in interrupt mode, meaning that statement (3) is not necessarily binding. Right? Then the specs are consistent. But if we treat the OR in statement (3) as an XOR, then the specs are inconsistent. Or is this assessment missing something? $\endgroup$
    – skytwosea
    Oct 17 at 21:18
  • $\begingroup$ (A) In Mathematical logic , we generally use regular OR , not XOR , @skytwosea , though that is irrelevant here. (B) SPECS are inconsistent here , which is easy to see via my Answer. (C) Your assessment is missing something ! (D) You have to show Consistency or InConsistency of the Whole Set , not a Single Statement like (3) while ignoring the other Statements. (E) You can try to assign your own truth values to those Criteria to make each Statement true & you will see that at least 1 Statement will be not true. $\endgroup$
    – Prem
    Oct 22 at 18:56

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