UGC NET CSE | January 2017 | Part 2 | Question: 6

Question

In propositional logic if $\left ( P \rightarrow Q \right )\wedge \left ( R \rightarrow S \right )$ and $\left ( P \vee R \right )$ are two premises such that

$$\begin{array}{c} (P \to Q) \wedge (R \to S) \\  P \vee R \\ \hline Y \\ \hline \end{array}$$

$Y$ is the premise :

• A. $P \vee R$
• B. $P \vee S$
• C. $Q \vee R$
• D. $Q \vee S$

Comments

Why $ Y can't \; be\;P\vee R$ ??(which is obvious implication from premises)

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PVR is premise not conclusion e.g if p and p->q is given  what is the conclusion q  right

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Answer will be A,B,C,D. All the options are Valid Conclusions from the given premises.

Find detailed answer here:

https://gateoverflow.in/335157/ugc-net-cse-january-2017-part-2-question-6?show=390553#a390553 

Answer 1

Answer: A, B,C, D

Each option is a Valid Conclusion of the given premises.

• A. $P \vee R$ trivially follows from $P \vee R.$

• A. $P \vee R$ is given. If $R$ then due to $R \rightarrow S,$ $S$ will follow, hence, $P \vee S$ will follow. If $P$ then $P \vee S$ will follow. So, in every case, $P \vee S$ will follow.

100. $P \vee R$ is given. If $P$ then due to $P \rightarrow Q,$ $Q$ will follow, hence, $Q \vee R$ will follow. If $R$ then $Q \vee R$ will follow. So, in every case, $Q \vee R$ will follow.

500. $P \vee R$ is given. If $P$ then due to $P \rightarrow Q,$ $Q$ will follow, hence, $Q \vee S$ will follow. If $R$ then due to $R \rightarrow S,$ $S$ will follow, hence, $Q \vee S$ will follow.

A nice question, But the answer will be All the options.

Comments

accidently a nice question but not intentionally made  . your analysis made it nice

Answer 2

Given that premises are

 (P→Q)˄(R→S)

 (P˅R) 

   (P→Q)   = ~PVQ

   (R→S)  = ~RVS

   (P˅R) 

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  Q V S

There will be Resolution (rule of inference ) between these premises to give conclusion  

~ P & P ,  R & R' will resolve out and then we  construct the disjunction of the remaining clauses

  to give SVQ option 4)

Comments

if $((P \rightarrow Q)\wedge(R\rightarrow S)) \wedge (P \vee R)\Rightarrow Y$

Then Y= (P v R) also satisfy it.

$\because$ the given formula is wrong only if LHS is true and rhs is false. Here if both premises true then LHS will be true so rhs should've to be true so (P or Q) is taken as Y for making rhs true.

Where i am doing wrong??

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Answer will be A,B,C,D. All the options are Valid Conclusions from the given premises.

Find detailed answer here:

https://gateoverflow.in/335157/ugc-net-cse-january-2017-part-2-question-6?show=390553#a390553 

Answer 3

One can think like -

For P-->Q: If you work hard(P) then you will qualify NET(Q).

For R-->S: If you play game(R) then you will get gold(S).

Given premise is PVR : means if u work hard(P) OR(v) if you play game(R), then conclusion will be-

than You will qualify NET(Q) OR(v) you will get gold(S). Which is nothing but QVS.

Hence, Option D is correct.

Answer 4

$(P\to Q)\wedge (R\to S)$  means if $P$ is true, then $Q$ has to be true, and if $R$ is true then $S$ has to be true.

The second statement $P \vee R$, says that either $P$ or $R$ is true, which means either $Q$ or $S$ should be true. 

Correct option is (D)  $Q\vee S$.