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The CORRECT formula for the sentence, "not all Rainy days are Cold" is

1. $\forall d (\text{Rainy}(d) \wedge \text{~Cold}(d))$
2. $\forall d ( \text{~Rainy}(d) \to \text{Cold}(d))$
3. $\exists d(\text{~Rainy}(d) \to \text{Cold}(d))$
4. $\exists d(\text{Rainy}(d) \wedge \text{~Cold}(d))$
edited | 1.3k views
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Is there typo in last option ? please correct it !
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corrected.
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You Choose UOD(Universe Of Discourse) Correctly , you can answer easily

Not all rainy days are cold.

In other words it says $\text{Some rainy days are not cold"}$

Given statement is
$\neg \forall d[R(d)\to C(d)]$
$\equiv \neg \forall d[\neg R(d) \vee C(d)]$
$\equiv \exists d[R(d)\wedge \neg C(d)]$
Hence option (D) is correct.

edited by
A) No rainy days are cold

B) All non-rainy days are cold

C)Some non-rainy days are cold.

D) Some rainy days are not cold.

option D
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Is option (A) statement correct?
+5
Statement A  shoud be "all days are rainy days and they are not cold "
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Now (A): "all days are rainy days and they are not cold " is the correct translation.
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The translation of option (C) should be,

(C) ∃d(~R(d)->C(d)) = ∃d(R(d) V C(d)) = (∃dR(d))  V (∃dC(d))  ="Some day are Rainy days or some days are Cold"

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option a statement may be , " all the rainy day are cold "

if i am wrong correct me please

Try this way

NOT (all rainy days are cold)

~($\forall$d Rainy(d)->Cold(d))

~($\forall$ ~Rainy(d) $\vee$cold(d))

$\exists$d( Rainy (d) $\wedge$~Cold(d))

OPTION D

edited
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Nicely explained

not all rainy days are cold : meaning "there are some rainy days which are cold" = "some days are rainy and not cold".

∃d{R(d)  ¬C(d)}

ans = option D

"all Rainy days are Cold" : ∀d(Rainy(d)->Cold(d))
"not all Rainy days are Cold" : ~∀d(Rainy(d)->Cold(d))
<=>∃d~(Rainy(d)->Cold(d))
<=>∃d~(~Rainy(d)VCold(d))
<=>∃d(Rainy(d)∧~Cold(d))

so Ans D is correct

(A)∀d(R(d)⋀~C(d)) = d(~(~R(d) V C(d))) (taking negation common)

=∀d(~(R(d)->C(d)))= All days are not Rainy days and also are not Cold

(B)d(~R(d)->C(d))=The day which are not Rainy day are Cold

(C)∃d(~R(d)->C(d))=∃d(R(d)VC(d))=Some day are Rainy days or some days are Cold

(D)∃d(R(d)⋀~C(d))= Some Rainy days are not Cold

= ~ (∀d(R(d)->C(d))) (taking negation common)

=not all Rainy days are Cold

edited by
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Is option (A)  Translation correct?

It should be "all days are rainy days and they are not cold ".

(A) Note that (p ∧ ~q) ≡ ~(p -> q). So it means rainy day to cold implication is false for all days. Which means non-rainy days are cold. (B) For all days, if day is not rainy, then it is cold [Non-Rainy days are cold] (C) There exist some days for which not rainy implies cold. [Some non-rainy days are cold] (D) Note that (p ∧ ~q) ≡ ~(p -> q). So it means rainy day to cold implication is false for some days. Which means not all rainy days are cold.

"Not all rainy days are cold."

Which means..

There is a rainy day which is not cold.

Which is equivalent to ∃d(Rainy(d)∧~Cold(d))

(as restriction of an existential quantification is same as existential quantification of a conjunction.)

so option D is correct.

"Not all Rainy days are Cold" are also rewrite “There exists at least one  Rainy day are  not Cold”

–1 vote

Option d bcz it means there is some days which are rainy not cold

+1
we get to conclusion

1.some rainy day are cold

or

2.some rainy day are not cold

so according to option d we get sme rainy day are not cold ..

so may be it is correct method .

if i am in correct then please correct me

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