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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 | 795 views
Is there typo in last option ? please correct it !
corrected.

You Choose UOD(Universe Of Discourse) Correctly , you can answer easily

Not all rainy days are cold.

In other words it says "Some rainy days are cold" or "Some rainy days are not cold"

Given statement is
~Vd[R(d)->C(d)]
<=>~Vd[~R(d)VC(d)]

<=> ∃ d[R(d) ^ ~C(d)]
D)

selected
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
Is option (A) statement correct?
Statement A  shoud be "all days are rainy days and they are not cold "
Now (A): "all days are rainy days and they are not cold " is the correct translation.

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"

Try this way

NOT (all rainy days are cold)

~(¥ d Rainy(d)->Cold(d))

~(¥d ~Rainy(d) DISJUNCTION cold(d))

∃d( Rainy (d) CONJUNCTION ~Cold(d))

OPTION D
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

(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
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.
"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

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