GO Classes CS Test Series 2025 | Discrete Mathematics | Topic Wise Test 1 | Question: 11

Answer 1

Detailed Video Solution: https://www.youtube.com/watch?v=WpgF9nv6Uxo&t=2336s 

Note that when an expression $\text{“A”}$ doesn’t have “$x$” as a free variable, then :



So, option A, C are equivalent. And B, D are equivalent.

“If someone is noisy, everybody is annoyed” $= \exists x\text{N}(x) \rightarrow \forall y\text{A}(y)$

So, B, D are true.

https://www.youtube.com/watch?v=WpgF9nv6Uxo&t=2336s 

Comments

Option A : ∃x(N(x) →∀y(A(y)))

There exists x, if x is noisy then everyone is annoyed.

I’m not able to find, why this is not equivalent with grammar given.
Why this option is wrong. I’m confused.

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∃x(N(x) →∀y(A(y)))

Consider the scenario: In a class, student A makes noise, student B doesn’t, and let’s assume No one is annoyed.

For this scenario, the statement “If someone is noisy, everybody is annoyed” becomes false, But option A becomes True.

∃x(N(x) →∀y(A(y))) basically is True if there is at least one person who is not making noise, regardless of whether everyone is annoayed or not.

Is the doubt clear? Comment if there any more doubt.

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 ∃x(P(x) → Q) ≡ ∃xP(x) → Q
also,
 ∀x(P(x) → Q) ≡ ∀xP(x) → Q

Is this valid? since Q does not have a free occurrence of x

EDIT: no, they are not equivalent.

Answer 2

B, D

Comments

Nice explanation!!

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Great method !