GO Classes CS 2025 | Weekly Quiz 1 | Propositional Logic | Question: 6

Answer 1

Given: $(p\,\wedge\sim q )\, \wedge(p\,\wedge r) \rightarrow (\sim p\,\vee q)$ is $\textbf{False}$

This is only possible when $(p\,\wedge\sim q )\, \wedge(p\,\wedge r)$ is $\textbf{True}$ and $(\sim p\,\vee q)$ is $\textbf{False}$.

1. $(\sim p\,\vee q)$ is $\textbf{False}$ only when $\sim p\, and\, q$ are $\textbf{False}$ which gives us $p = \textbf{True}$ and $q=\textbf{False}$.

2. Now, $(p\,\wedge\sim q )\, \wedge(p\,\wedge r)$ is $\textbf{True}$ and we know $p = \textbf{True}$ and $q=\textbf{False}$, so $r=\textbf{True}$

Hence, $\textbf{B}$ is the correct answer which is $p=\textbf{True}$, $q=\textbf{False}$ and $r=\textbf{True}$

Answer 2

The given expression is :(p $\wedge$ ~q) $\wedge$( p $\wedge$ r) → ( ~p $\vee$ q)
which is similar to (p $\wedge$ ~q) $\wedge$( p $\wedge$ r) → ( p →  q) [$\because$ p→ q is logically equivalent to ‘p+q]

This expression is claimed to be false 
Let consider $\alpha$ and $\beta$ as follows:

$\beta$ has to be false and $\alpha$ has to be true to make it false as a whole
For $\beta$ to be false p should be True and Q should be False
Now coming to $\alpha$ : (p $\wedge$ ~q) $\wedge$( p $\wedge$ r) After placing p as T and q as F we get T  $\wedge$ (T  $\wedge$ r) $\Rightarrow$ r
Inorder to make ​​​​​​ $\alpha$ true r has to assigned true

Correct answer will be option B .

Answer 3

There are multiple ways to solve this kind of questions, as others have used $Case \ Method$, so let me share my $Boolean \ Algebra$ approach. 

 

$(p \ $∧ ~ $q)$ $∧$ $(p ∧ r) →  $  ~ $p \ V \ q$

$= (pq’)(pr) \ → (p’ + q)$

$= pq’r → (p’ + q)$

$= (pq’r)’ + p’ + q$

$= p’ + q + r’ + p’ + q$

$= p’ + q + r’$

Now,  $p’ + q + r’$ is $False$ means $p’ , q\ , r’$ all must be $False$. [since OR operation]

1. e  $p’ = False ==> p= True$
2. $q = False$
3. $r’ = False ==> r= True$

 

$ So, p = True,  q = False, r = True$ 

 

Correct Ans: $Option \ B $

 

 

Answer 4

Given expression: $(p \wedge \neg q) \wedge (p \wedge r) \rightarrow \neg p \vee q$ 

Let’s assume, $\alpha$ = $(p \wedge \neg q) \wedge (p \wedge r)$;  $\beta$ = $\neg p \vee q$

Question is saying given expression is FALSE.

Implication is only FALSE for one condition $T \rightarrow F$. So, here $\alpha$ must be TRUE & $\beta$ must be FALSE.

So, $\beta$ = $\neg p \vee q$ = $F$ 

($\beta$ will be FALSE only if $\neg p$ and $q$ both are false.) [OR is only FALSE when both are false]

So, $\neg p$ = $F$; $q$ = $F$ 

Now we got, $p$ = $T$ and $q$ = $F$ 

Now $\alpha$ must be TRUE. 

So, $\alpha$ = $(p \wedge \neg q) \wedge (p \wedge r)$ = $T$ 

We can clearly see that it will be TRUE only if $(p \wedge \neg q)$ and $(p \wedge r)$ both are TRUE.

$(p \wedge \neg q) = (T \wedge \neg F) = (T \wedge T) = T$

$(p \wedge r) = (T \wedge r) = T$  ← This will be TRUE  only if r is TRUE. [AND  is TRUE only in one condition $T \wedge T = T$] 

So, $r$ = $T$.

Finally we got the truth values of $p, q, r$ as $T, F, T$ respectively.

Option (B) is correct.