GO Classes CS 2025 | Weekly Quiz 1 | Propositional Logic | Question: 7

Answer 1

Given p is true, so anything → p  $\equiv$ anything → true is true. Ans is A.

Answer 2

The given expression is $((~p’ $$\vee$$ q)$$\wedge$r$)→ p$ 
$p$ is T,$q$ is F and $r$ is T
The given expression after placing the truth values will evaluate to $Anything → True$ which is always True 
So whatever may be the value of q and r really doesn’t matter we just need $Anything → True$ or $ False → anything$ to claim the proposition to be true always or valid 

Correct answer will be option A,C,D

Answer 3

Given, ((~p V q) ∧ r ) → p  =((~T V F) ∧ T) → T  = F → T = T  [Option A]

 

if r=F,  ((~T V F) ∧ F) → T = F → T =T [Option C]

 

if q=T, ((~T V T) ∧ T) → T =((F V T) ∧ T) → T= T→ T = T [Option D]

 

Answer 4

$p$, $q$, $r$ are simple statements and truth values are $T, F, T$ respectively.

Now, $((\neg p \vee q) \wedge r) \rightarrow p$

$((\neg p \vee q) \wedge r) \rightarrow p\equiv$ $((F \vee F) \wedge T) \rightarrow T\equiv$ $(F \wedge T) \rightarrow T\equiv$ $F \rightarrow T\equiv$ $T$ 

So, truth value of given expression: $((\neg p \vee q) \wedge r) \rightarrow p$ is TRUE.

Also, Option (C.) and (D.) are correct here.

In option (C.) if $r$ is false then given expression is TRUE.

$((\neg p \vee q) \wedge r) \rightarrow p \equiv$ $((F \vee F) \wedge F) \rightarrow T\equiv$ $(F \wedge F) \rightarrow T\equiv$ $F \rightarrow T\equiv$ $T$ 

In option (D.) if $q$ is true then given expression is TRUE.

$((\neg p \vee q) \wedge r) \rightarrow p \equiv$ $((F \vee T) \wedge T) \rightarrow T\equiv$ $(T \wedge T) \rightarrow T\equiv$ $T \rightarrow T\equiv$ $T$ 

Ans. is: A;C;D

Comments

dumb method