2.3k 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. $Q\vee S$
| 2.3k 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

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)

by Boss (49.3k 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??
+1 vote

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.

by Active (2.1k points)