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.

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Null Quantification Rule Lecture (MUST WATCH): Null Quantification in Predicate Logic (Click HERE)

Questions & Variations based on Null Quantification Rule: https://www.youtube.com/watch?v=IoC5wmyf9ZQ&t=2407s (TimeStamp 00:40:07 Onwards)

Comments

 ∃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.

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@Deepak Poonia Sir, Please explain why in option A, ∃x(P(x) → ∀y(A(y))) != ∃xP(x) → ∀y(A(y)) 

As ∀y(A(y))) contains no free variable x?

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@____ , Watch the following lectures once.

Null Quantification Rule Lecture (MUST WATCH): Null Quantification in Predicate Logic (Click HERE)

Questions & Variations based on Null Quantification Rule: https://www.youtube.com/watch?v=IoC5wmyf9ZQ&t=2407s (TimeStamp 00:40:07 Onwards)

Answer 2

B, D

Comments

Nice explanation!!

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Great method !