The compound proposition which is always false is \[\]
A.$\left( p\to q \right)\leftrightarrow \left( \tilde{\ }q\to \tilde{\ }p \right)$\[\]
B.$\left[ \left( p\to q \right)\wedge \left( q\to r \right) \right]\leftrightarrow \left( p\to r \right)$\[\]
C. $\left( \tilde{\ }p\vee q \right)\leftrightarrow \left( p\to \tilde{\ }q \right)$\[\]
D. $p\to \tilde{\ }p$\[\]
Answer
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Hint: We use the definitions of negation $\tilde{\ }p$, conjunction $p\wedge q$, disjunction $p\vee q$, implication $p\to q$ and bi-implication $p\leftrightarrow q$ to find the truth values of compositions within the brackets if there is in each option and then the whole composite statement.
Complete step-by-step solution:
We know from the mathematical logic that if the statement $p$ has a truth value T or F then the negation of $p$ is denoted as $\tilde{\ }p$ and has truth value F or T respectively. The through the table of $\tilde{\ }p$ is given as
We also know that when there are two statements $p$ and $q$, the statement with the conjunction (with logical connective AND) of their truth values is denoted as $p\wedge q$ and has a truth value T only when both $p$ and $q$ have truth values, T, otherwise false. The truth table of $p\wedge q$ is
The statement with disjunction (with logical connective OR) of their truth values are denoted as $p\hat{\ }q$ and has a truth value T only when one of $p$ and $q$ have truth value T, otherwise false. The truth table of $p\vee q$ is
The statement with the implication (with logical connective If...then...) of their truth values is denoted as $p\to q$ and has a truth value F only when one of $p$ has a truth value T and $q$ has a truth value $F$ otherwise true. The truth table of $p\to q$ is
The statement with bi-implication (with logical connective if and only if ) of their truth values is denoted as $p\leftrightarrow q$ and has a truth value T only when both $p$ and $q$ have truth value T or truth value $F$, otherwise true. The truth table of $p\leftrightarrow q$ is
We use all these definitions to get the truth values in each option. We first find the truth values inside the bracket and then we find the truth values of whole composite statements. Let us denote the resultant statement in each option as $s$.\[\]
The truth table for option A: $s=\left( p\to q \right)\leftrightarrow \left( \tilde{\ }q\to \tilde{\ }p \right)$.
So option A is not correct. Let us move to option-B where three statements are involved . So let $t=\left( p\to q \right)\wedge \left( q\to r \right)$ , $s=\left[ \left( p\to q \right)\wedge \left( q\to r \right) \right]\to \left( p\to r \right)=t\to \left( p\to r \right)$ and draw the truth table .
So option B is not correct. We move to option C where $s=\left( \tilde{\ }p\vee q \right)\leftrightarrow \left( p\to \tilde{\ }q \right)$
So option C is correct . We move to option D where $s=p\to \tilde{\ }p$
So the only correct option is C. \[\]
Note: If the composite statement is always true then it is called a tautology and if the composite statement is always false it is called a fallacy. The statement $\tilde{\ }q\to \tilde{\ }p$ is the contrapositive of $p\to q$ and $q\to p$ is the converse of $p\to q$.
Complete step-by-step solution:
We know from the mathematical logic that if the statement $p$ has a truth value T or F then the negation of $p$ is denoted as $\tilde{\ }p$ and has truth value F or T respectively. The through the table of $\tilde{\ }p$ is given as
| $p$ | $\tilde{\ }p$ |
| T | F |
| F | T |
We also know that when there are two statements $p$ and $q$, the statement with the conjunction (with logical connective AND) of their truth values is denoted as $p\wedge q$ and has a truth value T only when both $p$ and $q$ have truth values, T, otherwise false. The truth table of $p\wedge q$ is
| $p$ | $q$ | $p\wedge q$ |
| T | T | T |
| T | F | F |
| F | T | F |
| F | F | F |
The statement with disjunction (with logical connective OR) of their truth values are denoted as $p\hat{\ }q$ and has a truth value T only when one of $p$ and $q$ have truth value T, otherwise false. The truth table of $p\vee q$ is
| $p$ | $q$ | $p\vee q$ |
| T | T | T |
| T | F | T |
| F | T | T |
| F | F | F |
The statement with the implication (with logical connective If...then...) of their truth values is denoted as $p\to q$ and has a truth value F only when one of $p$ has a truth value T and $q$ has a truth value $F$ otherwise true. The truth table of $p\to q$ is
| $p$ | $q$ | $p\to q$ |
| T | T | T |
| T | F | F |
| F | T | T |
| F | F | T |
The statement with bi-implication (with logical connective if and only if ) of their truth values is denoted as $p\leftrightarrow q$ and has a truth value T only when both $p$ and $q$ have truth value T or truth value $F$, otherwise true. The truth table of $p\leftrightarrow q$ is
| $p$ | $q$ | $p\leftrightarrow q$ |
| T | T | T |
| T | F | F |
| F | T | F |
| F | F | T |
We use all these definitions to get the truth values in each option. We first find the truth values inside the bracket and then we find the truth values of whole composite statements. Let us denote the resultant statement in each option as $s$.\[\]
The truth table for option A: $s=\left( p\to q \right)\leftrightarrow \left( \tilde{\ }q\to \tilde{\ }p \right)$.
| $p$ | $q$ | $\tilde{\ }p$ | $\tilde{\ }q$ | $p\to q$ | $\tilde{\ }q\to \tilde{\ }p$ | s |
| T | T | F | F | T | T | T |
| T | F | F | T | F | F | T |
| F | T | T | F | T | T | T |
| F | F | T | T | T | T | T |
So option A is not correct. Let us move to option-B where three statements are involved . So let $t=\left( p\to q \right)\wedge \left( q\to r \right)$ , $s=\left[ \left( p\to q \right)\wedge \left( q\to r \right) \right]\to \left( p\to r \right)=t\to \left( p\to r \right)$ and draw the truth table .
| $p$ | $q$ | $r$ | $p\to q$ | $q\to r$ | $p\to r$ | $t$ | $s$ |
| T | T | T | T | T | T | T | T |
| T | T | F | T | T | T | T | F |
| T | F | T | F | T | F | F | T |
| T | F | F | F | T | F | F | T |
| F | T | T | T | F | T | F | T |
| F | T | F | T | F | T | F | T |
| F | F | T | T | T | T | T | T |
| F | F | F | T | T | T | T | T |
So option B is not correct. We move to option C where $s=\left( \tilde{\ }p\vee q \right)\leftrightarrow \left( p\to \tilde{\ }q \right)$
| $p$ | $q$ | $\tilde{\ }p$ | $\tilde{\ }q$ | $\tilde{\ }p\vee q$ | $p\wedge \tilde{\ }q$ | s |
| T | T | F | F | T | F | F |
| T | F | F | T | F | T | F |
| F | T | T | F | T | F | F |
| F | F | T | T | T | F | F |
So option C is correct . We move to option D where $s=p\to \tilde{\ }p$
| $p$ | $\tilde{\ }p$ | $p\to \tilde{\ }p$ |
| T | F | T |
| F | T | F |
So the only correct option is C. \[\]
Note: If the composite statement is always true then it is called a tautology and if the composite statement is always false it is called a fallacy. The statement $\tilde{\ }q\to \tilde{\ }p$ is the contrapositive of $p\to q$ and $q\to p$ is the converse of $p\to q$.
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