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

Question

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?

Answer 1

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.

Answer 2

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

not all rainy days are cold:

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

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

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

Answer 3

(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.