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?

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    $\begingroup$ the statement can be paraphrased into "there is at least one rainy day that is not cold". $\endgroup$
    – jh4
    Feb 1 '15 at 23:22
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    $\begingroup$ Can you put each of the four options into English? i.e. What sentences to each of them represent? $\endgroup$ Feb 1 '15 at 23:26
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    $\begingroup$ Just my two cents: predicate logic symbols are obfuscating enough for statements about numbers, e.g., "not all odd numbers are prime" is much clearer than the same thing expressed with those predicate logic symbols. $\endgroup$ Feb 2 '15 at 2:44

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.

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    $\begingroup$ That's a very nice explanation. $\endgroup$
    – user174622
    Feb 1 '15 at 23:30
  • $\begingroup$ You need to correct last for Or and not AND I believe. Just a typo i think $\endgroup$
    Feb 1 '15 at 23:40
  • $\begingroup$ @ALEXANDER Thanks :) $\endgroup$ Feb 1 '15 at 23:44
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    $\begingroup$ @Sidsec9 I'm not exactly sure what you're asking here, but the equivalence $p\to q \equiv \neg p\lor q$ is pretty common (and helpful). Combined with DeMorgan, we see that $\neg(p\to q)\equiv \neg(\neg p\lor q) \equiv p\land \neg q$. $\endgroup$ Feb 2 '15 at 4:22
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    $\begingroup$ @Sidsec9 Note that another way to interpret the implication in "all rainy days are cold" is as a restricted universal quantifier, i.e. $\forall d\in R,Q(d)$ where $R$ is the set of all rainy days. This allows for a more direct reading as "for all rainy days, they are cold" rather than "for all days, if they are rainy then they are cold", and is clearly propositionally equivalent to induktio's interpretation given $d\in R\iff P(d)$. $\endgroup$ Feb 2 '15 at 11:50

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

not all rainy days are cold:

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

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

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

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    $\begingroup$ So it is a good answer. But I suggest to learn the latex here. $\endgroup$
    – peterh
    Feb 1 '15 at 23:27

(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.


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