How do you express system specifications using logical propositions p, q, and r?

  • Thread starter Thread starter VinnyCee
  • Start date Start date
  • Tags Tags
    Logic
VinnyCee
Messages
486
Reaction score
0

Homework Statement



Express the system specifications using the propositions p "The user enters a valid password," q "Access is granted," and r "The user has paid the subscription fee" and logical connectives.

a) The user has paid the subscription fee, but does not enter a valid password.

b) Access is granted whenever the user has paid the subscription fee and enters a valid password.

c) Access is denied if the user has not paid the subscription fee.

d) If the user has not entered a valid password but has paid the subscription fee, then access is granted.

Homework Equations



p = "The user enters a valid password"
q = "Access is granted"
r = "The user has paid the subscription fee"

The Attempt at a Solution



a) r\,\wedge\,\neg\,p

b) q\,\longleftrightarrow\,(r\,\wedge\,q)

c) \neg\,r\,\longrightarrow\,\neg\,q

d) (\neg\,p\,\wedge\,r)\,\longrightarrow\,qDo these answers look right?
 
Physics news on Phys.org
Everything looks right, except you made a typo (I guess) in b), it should be q\,\longleftrightarrow\,(r\,\wedge\,p), unless I'm missing something.
 
Even modulo typos, b is not right. The implication is not both ways. It does not assert that the only way to get access is with subscription and a valid password.
 
matt grime said:
Even modulo typos, b is not right. The implication is not both ways. It does not assert that the only way to get access is with subscription and a valid password.

Excuse my mistake, it should be (r \wedge p) \Rightarrow q.
 
So they are all right except that B is really (r\,\wedge\,p)\,\longrightarrow\,q?
 
VinnyCee said:
So they are all right except that B is really (r\,\wedge\,p)\,\longrightarrow\,q?

Yes, you already have two answers which implied that. :wink:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top