in Mathematical Logic edited by
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22 votes

A set of propositions is called a system specification.

System specification is consistent if they do not contain conflicting requirements that could be used to derive a contradiction.
When specifications are not consistent, there would be no way to develop a system that satisfies all specifications.
Consider the following System Specifications:

  1. “Whenever the system software is being upgraded, users cannot access the file system. If users can access the file system, then they can save new files. If users cannot save new files, then the system software is not being upgraded.”
  2. “The diagnostic message is stored in the buffer or it is retransmitted. The diagnostic message is not stored in the buffer. If the diagnostic message is stored in the buffer, then it is retransmitted.”

Which of the above system specifications are consistent?

  1. Only $1$
  2. Only $2$
  3. Both
  4. None
in Mathematical Logic edited by
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4 Comments

edited by

@this-is-geek-sassy

The definition of “Consistent” is given.

Consistent = Satisfiable.

A set of statements is Consistent if it is possible to make all of them true simultaneously. 

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@Deepak Poonia sir where we will find these questions??

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@anirudhkumar18 It didn’t understand your message.

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1 Answer

7 votes
7 votes
Basically this question asking that find the options which can make all the condition true simultaneous

Option 1:

Let whenever the system s/w is upgraded == U

user can access the file again = A

can save file  = S
Now Option 1 is saying that  U→ ~A ,A→ S, U→ S

let U = T then T→ ~A =True => ~A = True => A =False
A→ S => False → S = True => S can be t=True Or False
U→ S => T→ S =True => S=True
so option 1 is consistent
Option 2: Similarly that comes out to be consistent

3 Comments

U→ ~A and A → S doesn’t mean  U → S (by transitive property)
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These were there statement given but the question was asking just one think that is there anyone condition what make U->S

And this is true by taking U as True check out my answer
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@kickassakash 

I don’t think the transitivity you mentioned holds. 

$U \rightarrow \neg A$ will have the following diagram.


 

 

 

 

 

 

 

$A \rightarrow S$ will have the following diagram

 

 

 

 

 

 

 

 

 

 

out of $3$ possible diagram f $U\rightarrow \neg A$ and  $A \rightarrow S$ the following possibility will disregard  $U\rightarrow S$

 

 

 

 

 

 

 

 

 

 

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