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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. $\exists d(\text{Rainy}(d) \wedge \text{~Cold}(d))$
in Mathematical Logic by Veteran (100k points)
edited by | 1.7k views
0
Is there typo in last option ? please correct it !
0
corrected.
0

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

0

One more similar question asked in gate2014-

https://gateoverflow.in/769/gate2014-1-1

10 Answers

+27 votes
Best answer

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.

by Active (3.8k points)
edited by
+20 votes
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
by Active (2.6k points)
0
Is option (A) statement correct?
+5
Statement A  shoud be "all days are rainy days and they are not cold "
0
Now (A): "all days are rainy days and they are not cold " is the correct translation.
+2

 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"

0
option a statement may be , " all the rainy day are cold "

 

if i am wrong correct me please
+11 votes

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

by Loyal (8.4k points)
edited by
0
Nicely explained
+6 votes

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) \scriptstyle \wedge ¬C(d)}

ans = option D

by Boss (30.5k points)
+3 votes
"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
by Active (2.6k points)
+2 votes

(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

by Veteran (115k points)
edited by
0
Is option (A)  Translation correct?

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

(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.

by Loyal (9.5k points)
0 votes
"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.
by Boss (15.8k points)
0 votes

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

by Boss (47.6k points)
–1 vote
Not all rainy days are cold which is

$\neg(\forall d(\text{Rainy}(d)\rightarrow \text{Cold}(d) ))$

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

$(\exists \neg d(\neg\text{Rainy}(d) \lor \text{Cold}(d) ))$

$(\exists d(\text{Rainy}(d) \land \neg\text{Cold}(d) ))$

Option $D$
by Boss (11k points)
edited by
+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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