GO Classes CS Test Series 2025 | Discrete Mathematics | Topic Wise Test 1| Question: 12

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GO Classes CS Test Series 2025 | Discrete Mathematics | Topic Wise Test 1| Question: 12

GO Classes asked Apr 14, 2022 • retagged Apr 19 by Lakshman Bhaiya

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We define a new quantifier, uniqueness quantifier, the symbol of which is $\exists!.$

For any predicate $\text{P}$ and universe $\text{U}, \exists! x \text{P}(x)$ means there is exactly one element in the universe for which $\text{P}$ is true.

Which of the following statements is(are) Valid ?

$\exists!x \text{P}(x) \wedge \exists!x \text{Q}(x) \Rightarrow \exists!x (\text{P}(x) \wedge \text{Q}(x))$
$\exists!x (\text{P}(x) \wedge \text{Q}(x)) \Rightarrow \exists!x \text{P}(x) \wedge \exists!x \text{Q}(x)$
$\exists!x \text{P}(x) \vee \exists!x \text{Q}(x) \Rightarrow \exists!x (\text{P}(x) \vee \text{Q}(x))$
$\exists!x (\text{P}(x) \vee \text{Q}(x)) \Rightarrow \exists!x \text{P}(x) \vee \exists!x \text{Q}(x)$

I, II, IV
I, III
II, III, IV
IV only

GO Classes asked Apr 14, 2022 • retagged Apr 19 by Lakshman Bhaiya

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Deepak Poonia

commented Aug 31, 2022

i

edited

by Deepak Poonia Apr 21, 2023

Statement II is Invalid.

Assume the domain is set of all people in the world. Let $P(x)$ be “$x$ was Indian Prime Minister” ; Let $Q(x)$ be “$x$ was a Woman”

$\exists !x (P(x) \wedge Q(x))$ means that there is EXACTLY ONE person in the world who is satisfies $P(x) \wedge Q(x).$

So, this is True as we know (exactly one witness = Late Indira Gandhi).

But $\exists !x P(x) \wedge \exists !x Q(x)$ means that there is EXACTLY ONE person in the world who satisfies $P(x)$ (which is false because there are were many Indian Prime ministers) and there is Exactly One person in the world who satisfies $Q(x)$ (which is also false because there were many women).

So, II is Invalid.

$\exists !x (P(x) \wedge Q(x))$ means that there is EXACTLY ONE person in the world who is satisfies $P(x) \wedge Q(x).$

$\exists !x (P(x) \wedge Q(x))$ Does NOT mean that there is Exactly one person in the world who is satisfies $P(x)$ and there is Exactly one person in the world who is satisfies $Q(x).$

In simple terms, $\exists !x (P(x) \wedge Q(x)) \neq \exists !x P(x) \wedge \exists !x Q(x)$

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@Deepak Poonia sir, I have checked multiple time of option II (not option B) is getting the Valid . I could not make the counter example . Is II is valid? Please clarify? — diptanshu malviya, Aug 27, 2022
Statement II is Invalid.

Assume the domain is set of all people in the world. Let $P(x)$ be “$x$ was Indian Prime Minister” ; Let $Q(x)$ be “$x$ was a Woman”

$\exists !x (P(x) \wedge Q(x))$ means that there is EXACTLY ONE person in the world who is satisfies $P(x) \wedge Q(x).$

So, this is True as we know (exactly one witness = Late Indira Gandhi).

But $\exists !x P(x) \wedge \exists !x Q(x)$ means that there is EXACTLY ONE person in the world who satisfies $P(x)$ (which is false because there are were many Indian Prime ministers) and there is Exactly One person in the world who satisfies $Q(x)$ (which is also false because there were many women).

So, II is Invalid.
$\exists !x (P(x) \wedge Q(x))$ means that there is EXACTLY ONE person in the world who is satisfies $P(x) \wedge Q(x).$
$\exists !x (P(x) \wedge Q(x))$ Does NOT mean that there is Exactly one person in the world who is satisfies $P(x)$ and there is Exactly one person in the world who is satisfies $Q(x).$
In simple terms, $\exists !x (P(x) \wedge Q(x)) \neq \exists !x P(x) \wedge \exists !x Q(x)$ — Deepak Poonia, Aug 31, 2022
in option 1 , how can you make , LHS True, given, p is true and q is false — Nalinj, Apr 23
Please watch the Video Solution. We have proven that Statement $1$ is Not valid.
Statement 1:
LHS says that, there is EXACTLY ONE element $h$ in the domain for which $P(h)$ is true AND there is EXACTLY ONE element $j$ in the domain for which $Q(j)$ is true. (Note that $h,j$ may or may not be same element.)
RHS says that, there is EXACTLY ONE element $k$ in the domain for which $[P(k) \wedge Q(k)]$ is true.
Statement 1 is false & counter example is given HERE. — Deepak Poonia, May 5

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