Correct Answer - Option 2 :
\(\left( { \sim \left( {p \leftrightarrow q} \right) \wedge r} \right) \vee \left( {p \wedge q \wedge \sim r} \right)\)
Truth table: P → Q
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p
|
q
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p → q
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|
F
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F
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T
|
|
F
|
T
|
F
|
|
F
|
T
|
T
|
|
T
|
T
|
T
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Truth table: P ↔ Q
|
p
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q
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p ↔ q
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|
F
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F
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T
|
|
F
|
T
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F
|
|
T
|
F
|
F
|
|
T
|
T
|
T
|
Option 1
\(\left( {\left( {{\rm{p}} \leftrightarrow {\rm{q}}} \right) \wedge {\rm{r}}} \right) \vee \left( {{\rm{p}} \wedge {\rm{q}} \wedge \sim {\rm{r}}} \right)\)
\({\rm{Taking\;p}} = {\rm{T\;and\;q}} = {\rm{T}},{\rm{\;\;}}\left( {\left( {{\rm{T}} \leftrightarrow {\rm{T}}} \right) \wedge {\rm{r}}} \right) \vee \left( {{\rm{T}} \wedge {\rm{T}} \wedge \sim {\rm{r}}} \right)\)
\( = \left( {\left( {\rm{T}} \right) \wedge {\rm{r}}} \right) \vee \left( {{\rm{T}} \wedge \sim {\rm{r}}} \right)\)
\(= \left( {\rm{r}} \right) \vee \left( { \sim {\rm{r}}} \right)\)
\(= {\rm{T}}\)
The truth evaluation of Option_1 doesn’t depend on the value of r. If the value of r is True (T), then also this function will evaluate to T. Hence, it becomes the condition of AT LEAST TWO an fails for the condition of Exactly Two.
Option_2
\(\left( {\sim \left( {p \leftrightarrow q} \right) \wedge r} \right) \vee \left( {p \wedge q \wedge \sim r} \right)\)
\(= taking\;p = T\;and\;q = T,\;\;\left( {\sim \left( {T \leftrightarrow T} \right) \wedge r} \right) \vee \left( {T \wedge T \wedge \sim r} \right)\)
\(= \left( {\sim \left( T \right) \wedge r} \right) \vee \left( {T \wedge \sim r} \right)\)
\(= \left( {\sim\left( T \right) \wedge r} \right) \vee \left( {T \wedge \sim r} \right)\)
\( = \left( {F \wedge r} \right) \vee \left( {T \wedge \sim r} \right)\)
\(= \left( F \right) \vee \left( { \sim r} \right)\)
\(= \sim r\)
If r = F, then the output = T
else output = F.
Hence, Option_2 evaluates to T only when exactly p and q are True and r is necessarily False.
The same can be tried with keeping p and r as T OR q and r as T initially.
Option_3
\(\left( {\left( {p \to q} \right) \wedge r} \right) \vee \left( {p \wedge q \wedge \sim r} \right)\)
\(= taking\;p = T\;and\;q = T,\;\;\left( {\left( {T \to T} \right) \wedge r} \right) \vee \left( {T \wedge T \wedge \sim r} \right)\)
\(= \left( {\left( T \right) \wedge r} \right) \vee \left( {T \wedge \sim r} \right)\)
\(= \left( r \right) \vee \left( { \sim r} \right)\)
\(= T\)
The evaluation of final output does not depend on the value of r. Hence, the condition of EXACTLY TWO variables as T fails.
Option_4
\(\left( { \sim \left( {p \leftrightarrow q} \right) \wedge r} \right) \wedge \left( {p \wedge q \wedge \sim r} \right)\)
\(= taking\;p = T\;and\;q = T,\;\;\left( {\sim\left( {T \leftrightarrow T} \right) \wedge r} \right) \wedge \left( {T \wedge T \wedge \sim r} \right)\)
\(= \left( {\sim \left( T \right) \wedge r} \right) \wedge \left( {T \wedge \sim r} \right)\)
\(= \left( {F \wedge r} \right) \wedge \left( {T \wedge \sim r} \right)\)
\(= \left( F \right) \wedge \left( { \sim r} \right)\)
\(= F\)
The expression evaluates to F with r = T and also with r = F. Hence, with at least two variables as T, this evaluates to F.
