The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
+2 votes

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 in CBSE/UGC NET by Active (4k points) | 1.9k views

Given that 



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)

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



totally unclear question to me
Y is not premise it should be conclusion
Why $ Y can't \; be\;P\vee R$ ??(which is obvious implication from premises)
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)   = ~PVQ

   (R→S)  = ~RVS


  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 (50.5k points)
selected by
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??

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
47,919 questions
52,324 answers
67,778 users