GO Classes CS 2025 | Weekly Quiz 1 | Propositional Logic | Question: 9

Answer 1

$S$ is the sufficient condition for $P$ which mean if $S$ happens then $P$ will happen .In other words,$S→ P$
$R$ is a necessary condition for $P$ which means if $R$ doesn’t happen then $P$ doesn’t happen.This statement we can write as if $P$ happens then $R$ will happen OR $P→ R$

The two expressions we have are :

• S→ P
• P→ R

Now using the transitivity we can say $S→ P→ R$  $\Rightarrow$ $S→ R$

From this expression we can claim that S is the sufficient condition for R to happen and R is the necessary condition for S to happen 
Correct answer will be only option A.

Answer 2

$S$ is a sufficient condition for $P$, and $R$ is a necessary condition for $P$.
So, whenever $S$ is true, then $P$ is true. And whenever $P$ is true then $R$ is true. So, whenever $S$ is true, then $R$ is true(because when $S$ becomes true, then $P$ becomes true, then $R$ becomes true because $P$ has become true).
So, $S$ implies $R$. So, $S$ is a sufficient condition for $R$.
So, Whenever $S$ is true, $R$ is True.
So, sufficient condition for $P$ implies necessary condition for $P$.

Comments

when p is true, how is r true? r is a necessary condition for p right? so how does r depend on p? can you please explain?

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If p is true then r has to be true because p->r has to be true as given in the question

if p is true and if r will be false then overall whole p→ r will be false.

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@GO Classes What does “S is neither a sufficient nor a necessary condition for R” mean in Propositional logic terms, does it mean S Bi implication R is false?

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@nish97

since this is a multi-select question, its good to analyze all the options :

for option C : S neither sufficient nor a necessary condition for R can be expressed as →

= ((S is not sufficient for R) and (S is not necessary for R)) 

= (~(S → R) & ~(R → S))

= ~(S → R OR R → S)

= ~(~S + R + ~R + S )

= ~T 

= F

Note : (neither sufficient nor necessary) != ~(sufficient and necessary)      [~Bi-implication]

           (neither sufficient nor necessary) = (~sufficient & ~necessary) 

            you can further simplify it using de morgan's laws.

Answer 3

If $S → P$ and $P → R$ , then $S → R$.

This is called transitivity property or hypothetical syllogism. 

If $S → R$ then $S$ is a sufficient condition of $R$ and $R$ is a necessary condition for $S$.

Therefore $S$ is sufficient condition for $R$.

Answer $A$.

Comments

Very clear explanation.

Answer 4

according to statement

s->p

p->R

then , s->p->R

so  option A is correct .

Comments

Hypothetical Syllogism