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+23 votes

All that glitters is gold. No gold is silver.


  1. No silver glitters.
  2. Some gold glitters.

Then, which of the following is TRUE?

  1. Only claim $1$ follows.
  2. Only claim $2$ follows.
  3. Either claim $1$ or claim $2$ follows but not both.
  4. Neither claim $1$ nor claim $2$ follows.
  5. Both claim $1$ and claim $2$ follow.
in Mathematical Logic by Boss (30.8k points)
edited by | 1.1k views

4 Answers

+55 votes
Best answer

The correct answer is option (a) Only claim 1 follows.

$\text{Glitters}(x) \implies \text{ Gold}(x) \implies \lnot \text{ Silver}(x)$. Hence, Claim 1 follows. If something Glitters, it cannot be Silver.

For claim 2:

The set of things that Glitter could be empty.

We can still assert that All that Glitters is Gold, because nothing Glitters in the first place.

So, in the case when nothing Glitters, there is no Gold that Glitters. Glitters is still a subset of Gold, but there is no element in the subset Glitters.

by Boss (22.9k points)
edited by

ok thank u. Yes the correct 2nd claim be "Some gold may glitters " Am I right sir?

Yes. Had it been "Some gold may glitter", it would have been True.
Very eye catching solution give it 2 like. :p

Your logic is solid. Even TIFR paper setter made mistakes in This questions ! Snap From Official Key !

Thanks @pragya & @Akash for clearance.

Nice explanation.Thanks, Pragy Agarwal

nice explanation...
But if we solve with the help of Logical Venn Diagram, then (e) should be the answer.
@pragy for proving second claim

Given that all that glitters is gold is like

$\forall x (Glitters(x) \rightarrow Gold(x))$

and this implication can be true if all substances don't glitter and still they are gold or if all substances don't glitter and they are also not gold, that makes the implication true.

So, we cannot say some gold glitters. Am i right?
You are correct. good approach
ans would be e both the conclusion follow here.
+7 votes
$$\require{enclose}\enclose{circle}{\matrix{\\\rm is~Gold\\\enclose{circle}{\matrix{\\\rm Glitters\\\,}}\\\,}}\quad\enclose{circle}{\matrix{\\\rm Silver\\\,}}$$

Ans will be (e)

glitter is subset of gold, So some gold glitter and some not glitter, So (2) follows

silver is totally other set, because when something glitters, it must be gold,and it is given that no gold is silver.  So, glittering element can never be silver Here (1) follows
by Veteran (119k points)
Claim $1$ is correct.

For Claim $2$, what if the set $\rm Glitters$ is empty?
then there will be no concept of glitter, because if there is glitter , it only be gold. otherwise there will be 2 distinct sets, (1)gold (2)silver.
can someone plz provide me the predicate logic for second.

1- for all x ( such that x glitters -> gold).

2- what will be this.

@srestha: No. Glitters could be an empty set. And then, claim 2 will become False. Hence, the correct answer is a) Only Claim 1 follows.

then there will be no concept of glitter

No. A concept can exist even if there is nothing in the universe that can be categorized as that concept. There are no members of an empty set. That doesn't mean that an empty set can't exist.



All that Glitters is Gold $\equiv \forall x: \Bigl (\text{Glitters}(x) \implies \text{ Gold}(x)\; \Bigr )$

No Gold is Silver $\begin{align}
&\equiv \lnot \exists x: \Bigl ( \text{Gold}(x) \land \text{ Silver}(x) \Bigr )\\[1em]
&\equiv \forall x: \Bigl ( \text{Gold}(x) \implies \lnot \text{ Silver}(x) \Bigr )\\[1em]
&\equiv \underbrace{\forall x: \Bigl ( \text{Silver}(x) \implies \lnot \text{ Gold}(x) \Bigr )}_{\rm No~Silver~is~Gold.}


srestha selected one is correct :)

@Praggy Sir.

If glitters is an empty set, then how it is possible that All that glitters is gold.
srestha mam according to me and you ans is e but best ans say ans is a so which is right ans ?
One thing sure according to IBPS bank exam option e is correct science total depend upon logic if miss little bit  you will be wrong way
0 votes
Gold(X) = X is Gold

Glitter(X) = X glitters

Silver(X) = X is Silver

All that glitters is gold =>   Glitter(X) -> Gold(X)   => $\overline{Glitter(X)} \vee Gold(X)$  .... (1)

No gold is silver  =>          $\overline{Gold(X)} \wedge Silver(X)$ ... (2)

Now from (1) and (2) we got

                                                   $\overline{Glitter(X)} \vee Gold(X)$

                                                  $\overline{Gold(X)} \wedge Silver(X)$


                                                   $\overline{Glitter(X)} \wedge Silver(X)$            


So Option A ...
by Boss (12.1k points)
–3 votes

Lets simplify this by another example.

1.All scientists are Human.--------------------every x in universe if x is scientist then x is human.----∀x:(scientist(x)⟹ Human(x))

2.No  human is Dog.-------------------------¬∃x:(human(x)∧ Dog(x))


1.No dog is scientist.----------For this to be false ,some dog should be scientist but scientist if exist are only humans and no human is dog hence above statement is true.

2.some human are scientist.----every scientist if exists is human,but if there are no scientists .then no human is scientist.Hence this is false.


by Active (4.8k points)

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