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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.

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.

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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