+2 votes
2.1k views

In propositional logic, if $(P→Q)\wedge (R→S)$ and $(P\vee R)$ are two premises such that $Y$ is the premise:

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

1. $P\vee R$
2. $P\vee S$
3. $Q\vee R$
4. $Q\vee S$
asked | 2.1k views
+1

Given that

(P→Q)˄(R→S)

(P˅R)

here if P then Q and if R then S  now P V R means either Q is true or S is true

so Y will be Q V S so ans is 4)

0
plz explain a little.. i know propositions. but i could not understand what the question is saying.. what are the operators between

(P→Q)˄(R→S)

(P˅R)

totally unclear question to me
0
Y is not premise it should be conclusion
0
Why $Y can't \; be\;P\vee R$ ??(which is obvious implication from premises)
+1
PVR is premise not conclusion e.g if p and p->q is given  what is the conclusion q  right

## 1 Answer

+3 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)

answered by Veteran (50.7k points)
selected by
0
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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