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April 24, 2014

April 24, 2014

Posted by **Jerry** on Wednesday, August 1, 2012 at 2:28am.

- Discrete Math -
**MathMate**, Wednesday, August 1, 2012 at 8:00amA nice puzzle.

The truth table for ~R is

R ~R

T F

F T

Since ~R=true, therefore R=false.

Given R is false, and

R v(P v T)=R ∨ P ∨ T is true, we conclude that P∨T is true, i.e.

**P=true ∨ T=true**.... (1)

Then we need the table for the conditional (->)

A B A->B

T T T

T F**F**

F T T

F F T

This means that the conditional is false if and only if the first proposition (A) is true and the second (B) is false.

Consequently, given that

~R->(S->T) is true

and that ~R=true, we conclude that

S->T = true .....(2)

We are also given that

P->Q = true ....(3)

From (1), we have two cases, either P is true or T is true.

case 1. P=true

From (3), P->Q = true, so if P is true, Q must be true.

case 2: T=true

(2) does not guarantee that S is true. So S can be true or false.

Hence, we do not know if (Q∨S) is true or false.

Can you check the question to see if there was a typo, such as

R v(P v T) was meant to be R∨(P∧T) ?

- Discrete Math -
**Jerry**, Wednesday, August 1, 2012 at 1:40pmYou came up with everything I came up with. I, too, believe something must be incorrect. I met with my professor today, and showed him my 25+ pages of attempts at solving it. He didn't want to go through the whole thing, but said it had to do with focusing on showing (QvS) could not be False (Q and S both being False). I didn't press him any further because I was still getting an A out of the course, but I'll probably still work on the problem to satisfy my own curiosity. Will try to post if I solve it.

- Discrete Math -
**MathMate**, Wednesday, August 1, 2012 at 3:51pmSorry, I came up with the same conclusion:

when P=false, T=true, we cannot determine the truth value of neither Q nor S.

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