GATE CSE 2017 Set 2 | Question: 11

Question

Let $p, q, r$ denote the statements ”It is raining”, “It is cold”, and “It is pleasant”, respectively. Then the statement “It is not raining and it is pleasant, and it is not pleasant only if it is raining and it is cold” is represented by

• A. $(\neg p \wedge r) \wedge (\neg r \rightarrow (p \wedge q))$
• B. $(\neg p \wedge r) \wedge ((p \wedge q) \rightarrow  \neg r)$
• C. $(\neg p \wedge r) \vee ((p \wedge q) \rightarrow  \neg r)$
• D. $(\neg p \wedge r) \vee (r \rightarrow (p \wedge q))$

Comments

as @vintl mentioned the reason, I too also feel the answer should be (c)

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https://gateoverflow.in/1642/gate1998-1-5?show=295562#c295562

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q only if p $=$ if no p then no q $=$ ~p → ~q $=$ q → p

Answer 1

1. "It is not raining and it is pleasant" can be written as $(¬p∧r)$

2. Now, "it is not pleasant only if it is raining and it is cold" is represented by $¬r\implies (p∧q)$  but $(p∧q) \not\implies ¬r $. Why? Because if it is not pleasant then we can conclude it must be raining and it is cold. However, it is raining and cold does not assure that it will be unpleasant. i.e., $p$ only if $q$ can be written as if $p$ then $q$ (not double implication).

So, ANDing clause $1.$ and $2.$ we get $(¬p∧r)∧(¬r→(p∧q))$

option A is correct.

Comments

So, ANDing clause 1. and 2. we have (¬p∧r)∧(¬r→(p∧q))(¬p∧r)∧(¬r→(p∧q)) [option A is correct].

I think some small correction ( *(¬p∧r)∧(¬r→(p∧q)) == (¬p∧r)∧(¬r→(p∧q)) ) is required in this line. 

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If there is a confusion b/w "only if" and "if" hopefully the following will clear that

It is a common mistake to read only if as a stronger form of if. It is important to emphasize that "q if p" means that p is a sufficient condition for q, and that "q only if p" means that p is a necessary condition for q.

 

Furthermore, we can supply more intuition on this fact: Consider q only if p. It means that q can occur only when p has occurred: so if we don't have p, we can't have q, because p is necessary for q. We note that if we don't have p, then we can't have q is a logical statement in itself: ¬p⇒¬q. We know that all logical statements of this form are equivalent to their contrapositives. Take the contrapositive of ¬p⇒¬q: it is ¬¬q⇒¬¬p, which is equivalent to q⇒p.

Ref: https://math.stackexchange.com/questions/617562/conditional-statements-only-if

This example might help:

"X kills blackbuck only if he has a gun with himself".

Q: X kills blackbuck

P: He has a gun with himself

P is a necessary condition for Q i.e. Having a gun with him is necessary to kill blackbuck.

If P is False then Q can't occur: If he doesn't have gun then he can't kill blackbuck.

But the mistake we usually do is we interpret the statement as P->Q which means that: 

1.If X has gun with himself then he kills blackbuck (i.e. if P=T then Q=T).

2.If X doesn't have gun with himself then we don't know what X does (i.e. if P=F then Q=?). Whether he buys/borrows and then kills or just leaves it.

But according to the statement we know that X will kill only if he has a gun with himself which means he has to have a gun already with himself to kill.

If he does not have a gun(P=F)  and since he does not buys/borrows it so he won't kill blackbuck i.e. Q=F. So if P' then Q' i.e. P'-> Q'  which is Q->P.

 

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@MiNiPanda

q if p" means that p is a sufficient condition for q, and that "q only if p" means that p is a necessary condition for q.

For both of this 1) q if p and 2) q only if p

q ==> p and $\widetilde{p}$  ==> q valid ??

But not p ==> q

correct me if wrong

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@jatin khachane 1,

1) q if p means p $\rightarrow$ q

2) q only if p means ~p $\rightarrow$ ~q which is q $\rightarrow$ p

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@jatin khachane 1,

Your interpretation has a slight mistake,

q if p.. means p is sufficient condition of q....but q can happen without p

 Upto this it is right. Then you said p $\rightarrow$ q is not valid. But why? Since q can happen without p so even if p is False, q can be anything (it may happen or not happen i.e. q can be T or F) and still the outcome should be valid. p $\rightarrow$ q exactly satisfies this condition. When p=F, q can be anything. And when p is T, q has to be T.

Eg: For checking whether a graph is Hamiltonian, Dirac's theorem is sufficient. Means if theorem is satisfied then graph is Hamiltonian but if not satisfied, then graph can be anything. We can't conclude.

p: Graph satisfies Dirac's Theorem.

q: Graph is Hamiltonian

p is sufficient condition of q

p $\rightarrow$ q

So when p is T, q has to be T. 

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@MiNiPanda

Yes bro ..I found that ..that y I hidden that comment :)

simply 

[ p sufficent for q ] , same as ,  [ q if p ]  ,same as,  [ if p then q ] ,  [p==>q]

[ p necessary for q ] , same as ,  [ q only if p ]  ,same as,  [ if ~p then ~q ] ,  [q==>p]

Now ok bro ?

 

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Can we think like this? :

¬r⟹(p∧q)

Here  ¬r  is false , so false implies anything is always true. Hence above expression will be TRUE.

But (p∧q)⟹¬r  in this expression  (p∧q) is TRUE and ¬r is FALSE , so it will be FALSE.Hence it is wrong.

Answer 2

it is not pleasant only if it is raining and it is cold

it is not pleasant ---> ¬r

t is raining and it is cold----> (p∧ q)

therefore it becomes  (¬r→(p∧q))

Comments

Which book ?

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Kenneth H.Rosen

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thank you for this :)

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helpful

Answer 3

$(\sim p \wedge r) \wedge (\sim r \rightarrow (p \wedge q))$

Answer is A

Answer 4

option A seems to be correct.