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The first order logic statement $((\text{R} \vee \text{Q}) \wedge (\text{P} \; \vee \sim \text{Q}))$ is equivalent to which of the following?

  1. $((\text{R} \vee \sim \text{Q}) \wedge (\text{P} \; \vee \sim \text{Q}) \wedge (\text{R} \vee \text{P}))$
  2. $((\text{R} \vee \text{Q}) \wedge (\text{P} \; \vee \sim \text{Q}) \wedge (\text{R} \vee \text{P}))$
  3. $((\text{R} \vee  \text{Q}) \wedge (\text{P} \; \vee \sim \text{Q}) \wedge (\text{R} \vee \sim \text{P}))$
  4. $((\text{R} \vee  \text{Q}) \wedge (\text{P} \; \vee \sim \text{Q}) \wedge ( \sim \text{R} \vee \text{P}))$
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Answer: C

We know that $a \rightarrow b = \lnot a \lor b$

So, $((R∨Q)∧(P∨∼Q))= ((\lnot R \rightarrow Q) \land (Q \rightarrow P))$, if this is true then

both $(\lnot R \rightarrow Q)$ and $(Q \rightarrow P)$ must be true and from this we can conclude that $(\lnot R \rightarrow P)$

We can write $(\lnot R \rightarrow P) = (R \lor \lnot P)$

Now we know that when $A \land B \rightarrow C$ so if $A \land B$ is true so $C$ must be true and when $A \land B$ evaulates to false we don’t care about truth value of $C$ because expression will ultimately evaluates to false.


So, $((R∨Q)∧(P∨∼Q))$ and $((R∨Q)∧(P∨∼Q) \land (R \lor \lnot P))$ will have same truth table hence both of these expression is equivalent.

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