in Mathematical Logic edited by
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Consider the statement

 "Not all that glitters is gold”

Predicate glitters$(x)$ is true if $x$ glitters and predicate gold$(x)$ is true if $x$ is gold.  Which one of the following logical formulae represents the above statement?

  1. $\forall x: \text{glitters} (x)\Rightarrow \neg \text{gold}(x)$
  2. $\forall x:\text{gold} (x)\Rightarrow \text{glitters}(x)$
  3. $\exists x: \text{gold}(x)\wedge \neg \text{glitters}(x)$
  4. $\exists x: \text{glitters}(x)\wedge \neg \text{gold}(x)$
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3 Comments

option B is not true because it says that " All gold glitters " which cannot be extracted from "Not all that glitters is gold"

we can only extract from this statement is "Some that glitters is not gold  "
We don't know about the gold that all of it glitters or not
2

Understand the Difference between this & this.

If it says "None of that glitters is gold" then $\sim (\forall x(glitters(x)\rightarrow Gold(x)))$ will not work because this is saying some of that glitters is not gold (or) all of that glitters is not gold,  not None of that glitters is gold.

1

Think domain as all the metals in the world.

Note :- glitter means “shines”

Understand meaning of each option in english below :-

Option A :- All the metals which glitter are not gold. (false because word “all” means gold are also included but gold always glitter)

Option B :- All the metals which are gold are glitter (although this statement is true but it not saying same thing as this statement “Not all that glitters is gold”)

Option C :- there exist a metal x which is gold but not glitter (this statement is always false)

Option D :- there exist a metal x which glitter but not gold (this is same statement that we have given in question)

So, option D is correct.

2

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4 Answers

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Best answer

"Not all that glitters is gold”  can be expressed as : 

$\neg(\forall x(\text{glitters}(x)\implies \text{gold}(x)))$ 

(as restriction of universal quantification is same as universal quantification of a conditional statement.)

"Not all that glitters is gold" means "some glitters are not gold" which can be expressed as 

 $\exists x(\text{glitters}(x)\wedge  \neg \text{gold}(x))$

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

So option (D) is correct. 

edited by

2 Comments

moved by

 "Not all that glitters is gold” 

6

i guess according to ur diagram , option b also becomes true!!!

for any metal,if it is gold=> it glitters

0
15 votes

Option D is correct .
              "Not all that glitters is gold
can be expressed as :


⇒∼(∀x(glitters(x)⇒gold(x))

⇒$ ∃x\neg(glitters(x)⇒gold(x))$

⇒$∃x(\neg(\neg glitters(x) \vee gold(x))$

⇒$∃x(glitters(x) \wedge \neg gold(x))$

can be expressed as :

         " some glitters are not gold"

 

edited
by

6 Comments

~(x(glitters(x)gold(x))

u have started from above why don't u use "^" between glitters(x),gold(x))

0
I think this will help you

“For every person x, if person x is a student in this class then x has studied calculus.”
If S(x) represents the statement that person x is in this class, we see that our statement can be
expressed as ∀x(S(x) → C(x)). [Caution! Our statement cannot be expressed as ∀x(S(x) ∧
C(x)) because this statement says that all people are students in this class and have studied
calculus!]
0
i have read this statement in kenneth rosen ..but in this question .. we have to think like this that expect gold some other thing can also glitters that why we dont use "AND" here..??
0

x(glitters(x)^gold(x) I think it's meaning is all things are glitters and it is gold.They said to represent all that glitters is gold

0
yaar sometimes mathematical logic seems to be difficult ...

??
0
0
1 vote
answerv is d
by
0 votes
The statement “Not all that glitters is gold” can be expressed as follows :

¬(∀x(glitters(x)⇒gold(x)) ... (1)

Where ∀x(glitters(x)⇒gold(x) refers that all glitters is gold. Now ,

∃x¬(glitters(x)⇒gold(x)) ... (2) , Since we know ¬∀x() = ∃x¬()

(Where ∀ refers to -> All and ∃x refers to -> There exists some).

As we know, A⇒B is true only in the case that either A is false or B is true. It can also defined in the other way :

A⇒B=¬A∨B (negationA or B ) ... (3)

From equation (2) and (3) , we have ∃x(¬(¬glitters(x)∨gold(x))

⇒∃x(glitters(x)∧¬gold(x)) ... (4) , Negation cancellation ¬(¬) = () : and ¬(()∨()) = (¬()∧¬()) .

So Answer is (D) .

1 comment

Complex answer ...!!
0
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