ans is q v s

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2 votes

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 :

- $P \vee R$
- $P \vee S$
- $Q \vee R$
- $Q \vee S$

4 votes

Best answer

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??

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??

0

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.