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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 | 626 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 "

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

(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.