in Mathematical Logic edited by
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44 votes
44 votes

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

  1. $(\neg p \wedge r) \wedge (\neg r \rightarrow (p \wedge q))$
  2. $(\neg p \wedge r) \wedge ((p \wedge q) \rightarrow  \neg r)$
  3. $(\neg p \wedge r) \vee ((p \wedge q) \rightarrow  \neg r)$
  4. $(\neg p \wedge r) \vee (r \rightarrow (p \wedge q))$
in Mathematical Logic edited by
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4 Comments

as @vintl mentioned the reason, I too also feel the answer should be (c)
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q only if p $=$ if no p then no q $=$ ~p → ~q $=$ q → p
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9 Answers

47 votes
47 votes
Best answer
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.
edited by

4 Comments

@

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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@

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.
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26 votes
26 votes

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))

4 Comments

Which book ?
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Kenneth H.Rosen
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thank you for this :)
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helpful
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7 votes
7 votes
$(\sim p \wedge r) \wedge (\sim r \rightarrow (p \wedge q))$

Answer is A
4 votes
4 votes
option A seems to be correct.
Answer:

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