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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 :

  1. $P \vee R$
  2. $P \vee S$
  3. $Q \vee R$
  4. $Q \vee S$
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4 Comments

It is one of the standard rule of inference(constructuctive dillema )

ans is q v s
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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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5 Answers

4 votes
4 votes
Best answer

Given that premises are

 (P→Q)˄(R→S)

 (P˅R) 

   (P→Q)   = ~PVQ

   (R→S)  = ~RVS

   (P˅R) 


  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)

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1 comment

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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2 votes
2 votes

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.

1 vote
1 vote

$(P\to Q)\wedge (R\to S)$  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)  $Q\vee S$.

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0 votes
Ans : D

 

Q∨S

This is based upon  constrctive dillema  rule of inference.
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