How to express the statement “not all rainy days are cold” using predicate logic?

I am trying to figure out how to express the sentence “not all rainy days are cold” using predicate logic. This is actually a multiple-choice exercise where the choices are as follows:

(A) $\forall d(\mathrm{Rainy}(d)\land \neg\mathrm{Cold}(d))$

(B) $\forall d(\neg\mathrm{Rainy}(d)\to \mathrm{Cold}(d))$

(C) $\exists d(\neg\mathrm{Rainy}(d)\to\mathrm{Cold}(d))$

(D) $\exists d(\mathrm{Rainy}(d)\land \neg\mathrm{Cold}(d))$

I am really having a hard time understanding how to read sentences correctly when they are in predicate logic notation. Can someone give me a hint on how to do this and also how to approach the problem above?

Solutions Collecting From Web of "How to express the statement “not all rainy days are cold” using predicate logic?"

Think about the positive statement first (of which your statement is the negation). That is, consider the following statement: “All rainy days are cold.”

Use the following notation:

$P(d):$ The day is rainy.

$Q(d):$ The day is cold.

Thus, we may represent the positive statement as follows:
$$
\forall d(P(d)\to Q(d)).\tag{1}
$$
The statement you are considering is the negation of $(1)$; that is, you are considering the statement, “Not all rainy days are cold.” Thus, you need to negate $(1)$:
$$
\neg[\forall d(P(d)\to Q(d))] \equiv \exists d\neg[P(d)\to Q(d)]\equiv \exists d[P(d)\land \neg Q(d)].
$$
Thus, the answer to your question is D.

The Answer would be D.. Hold on I’ll explain why..

not all rainy days are cold:

$\sim (\forall \text{d} (\text{Rainy}(\text{d}) \to \text{Cold}(\text{d})))$

$\equiv \sim(\forall \text{d}(\sim \text{Rainy}(\text{d}) \lor \text{Cold}(\text{d})))$

$\equiv \exists \text{d}(\text{Rainy}(\text{d}) \wedge \sim \text{Cold}(\text{d}))$

(A) All days are rainy and are not cold.

(B) All days are rainy, cold, or both. (Alternatively: All days that are not rainy are cold.)

(C) There is a day that is rainy, cold, or both.

(D) There is a day that is rainy and is not cold.

A and D are straightforward.

B and C are easy to misunderstand because the material implication operator is easy to misunderstand. The only time ~Rainy(d)→Cold(d) is false is when ~Rainy(d) is true and Cold(d) is false, that is, when Rainy(d) and Cold(d) are both false.

B is true when ~Rainy(d)→Cold(d) is true of all days; each day is rainy, cold, or both.

C is true when ~Rainy(d)→Cold(d) is true of some day d; that particular days is rainy, cold, or both.