GO Classes CS 2025 | Weekly Quiz 1 | Propositional Logic | Question: 10

Answer 1

$\begin{array}{|l|l|l|l|} \hline {} \text{X} & \text{Y }& \text{X\$Y }\\ \hline \text{1} & \text{0 }& \text{1 }\\ \hline \text{1} & \text{1}& \text{1 }\\ \hline \text{0} & \text{1 }& \text{0 }\\ \hline \text{0} & \text{0 }& \text{1 }\\ \hline \end{array}$

From the table, we can see that it is resembling the truth table of the implication operator for $Y\rightarrow X$. 

or,

We can also get the same result by writing the expression for truth values 1.

The expression for $X\$ Y$ is:

$=$ $X\neg Y \vee XY \vee \neg X \neg Y$

On simplifying we get,

$=$ $\neg Y \vee X$

and, $\neg Y \vee X\equiv Y \rightarrow X$

therefore, $X \$ Y$ $=$ $Y\rightarrow X$

A. $X \$ \neg Y \equiv \neg Y\rightarrow X$ which is not same as the $Y\rightarrow X$.

B. $\neg X \$ Y \equiv Y\rightarrow \neg X$ which is not same as the $Y\rightarrow X$.

C. $\neg X \$ \neg Y \equiv \neg Y \rightarrow \neg X$ which is not same as the $Y\rightarrow X$.

D. None of the above

Therefore, D is the correct option.

Comments

Can you tell me whatis $ in opetions ?

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This is not a standard operator, it is given in the question and you have to determine its working.

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This answer is perfectly alright but we can stop it at (xv~Y) = X+~Y then we can check all options for this also

Answer 2

 

Thanks you!!

Answer 3

• Hint : observe the truth table it is matching with which operator !!

  3 times true and 1 times false matching with the result of Implication operator, OR operator now who implies whom we can find it easily by observing that “for which one when it is false answer we are getting is true always and then check when it is true does then if another prop’s value is false then is it false then we can conclude that it implies to another prop"
  or We can just try to match the truth table of X→Y and truth table of Y→X and confirm which one matching with this truth table.
  ” what or what “is unclear so to FIND that we need to create an compound prop regarding that when this table give output as true . X¬Y ∨ XY ∨ ¬X¬Y

• = X(¬Y∨Y) ∨ ¬X¬Y

  = X ∨ (¬X¬Y)

  = (X v ¬X) and (X v ¬Y)=(X v ¬Y)

  = Y→X so Y→X== X$Y
  
  OPTION A . Is equivalent to not of y implies x not matching with actual compound prop
  OPTION B . is equivalent to Y implies not of X not matching
  OPTION C. is equivalent to not of Y implies not of X not matching
  HENCE OPTION D IS CORRECT