test series

Question

Which of the following arguments is invalid?

• A. $P \vee Q, \sim P \rightarrow R, Q \rightarrow S \vdash R \wedge S$
• B. $P \rightarrow \sim Q, R \rightarrow Q, R \vdash \sim P$
• C. $P \rightarrow R, Q \rightarrow R, Q \vee P \vdash R$
• D. $P \rightarrow \sim Q, \sim Q \vdash P$

Answer 1

A and D is invalid.

Answer 2

Option (a) is invalid. Given propositions $P \vee Q,{P}' \rightarrow R, Q \rightarrow S$ can be re-written as ${P}' \rightarrow Q, Q \rightarrow S, {P}' \rightarrow R$. ${P}' \rightarrow Q, Q \rightarrow S$ together gives ${P}' \rightarrow S$ by transitivity of implication. So the given propositions are now ${P}' \rightarrow S, {P}' \rightarrow R$. But this cannot mean $R \wedge S$, instead, it is equivalent to $R \vee S$ by constructive dilemma rule.

Option (b) is valid. $P\rightarrow {Q}',R\rightarrow Q, R$ can be re-written as ${P}' \vee {Q}',{R}' \vee Q,R$. Now,  ${R}' \vee Q,R$ is equivalent to $Q$. So effectively it reduces to ${P}' \vee {Q}',Q$ which is equivalent to ${P}'$ by Modus Tollens.

Option (c) is valid because the given set of propositions can be written as $P \rightarrow R, Q \rightarrow R, P, Q$ by applying simplification rule on $Q \vee P$. Then applying Modus Ponens, $P \rightarrow R, P, Q \rightarrow R, Q$ gives $R$.

Option (d) is valid. $P\rightarrow {Q}',{Q}'$ is equivalent to ${P}'\vee {Q}',{Q}'$which always produces $P$.

Therfore option (a) is invalid.

Comments

For option C, you have used simplification...

P ^ Q  => P or P ^ Q => Q is simplification rule right??? We should have ^ symbol....