
Which of the following is correct?
A. $(p \vee q) \equiv (p \wedge q)$
B. $(p \to q) \equiv (q \to p)$
C. $(p \to q) \equiv (p \wedge q)$
D. None of these
Answer
558.3k+ views
Hint: These questions can be solved by understanding the meaning and implications of the logical connectives like $\wedge$ , $\vee$ or $\to$ . Further, a truth table can be drawn for the given compound statements, and then each option can be evaluated and checked.
Complete step-by-step solution:
First, draw the truth table for each of the given compound statements in the question.
Truth Table for $(p \vee q)$ :
$(p \vee q)$ is true when either $p$ is true or $q$ is true or both are true. It is only false if both are false, simultaneously.
Truth Table for $(p \wedge q)$ :
$(p \wedge q)$ is true only when both, $p$ and $q$ are true, otherwise, it is always false.
Truth Table for $(p \to q)$ :
$(p \to q)$ is true if the condition while implying $q$ is true, otherwise, the output will be false. This is a two-valued logic.
Truth Table for $(q \to p)$ :
The same conditions are true for $(q \to p)$ logical statements, which are true for $(p \to q)$ logical statements.
Now, compare all the truth tables one by one according to the options given in the question.
Comparing all the truth tables, it is clear that the truth table for the logical statement $(p \to q)$ is equivalent to the truth table for the logical statement $(q \to p)$ .
Hence, the correct answer out of the given options is B. $(p \to q) \equiv (q \to p)$.
Note: Always be careful while writing the truth tables for logical statements like, $(p \wedge q)$ and $(p \vee q)$. Also, remember to carefully check each and every condition of the two truth tables sequentially to prove them equivalent to each other.
Complete step-by-step solution:
First, draw the truth table for each of the given compound statements in the question.
Truth Table for $(p \vee q)$ :
| $p$ | $q$ | $(p \vee q)$ |
| T | T | T |
| T | F | F |
| F | T | T |
| F | F | T |
$(p \vee q)$ is true when either $p$ is true or $q$ is true or both are true. It is only false if both are false, simultaneously.
Truth Table for $(p \wedge q)$ :
| $p$ | $q$ | $(p \wedge q)$ |
| T | T | T |
| T | F | F |
| F | T | F |
| F | F | F |
$(p \wedge q)$ is true only when both, $p$ and $q$ are true, otherwise, it is always false.
Truth Table for $(p \to q)$ :
| $p$ | $q$ | $(p \to q)$ |
| T | T | T |
| T | F | F |
| F | T | T |
| F | F | T |
$(p \to q)$ is true if the condition while implying $q$ is true, otherwise, the output will be false. This is a two-valued logic.
Truth Table for $(q \to p)$ :
| $q$ | $p$ | $(q \to p)$ |
| T | T | T |
| T | F | F |
| F | T | T |
| F | F | T |
The same conditions are true for $(q \to p)$ logical statements, which are true for $(p \to q)$ logical statements.
Now, compare all the truth tables one by one according to the options given in the question.
Comparing all the truth tables, it is clear that the truth table for the logical statement $(p \to q)$ is equivalent to the truth table for the logical statement $(q \to p)$ .
Hence, the correct answer out of the given options is B. $(p \to q) \equiv (q \to p)$.
Note: Always be careful while writing the truth tables for logical statements like, $(p \wedge q)$ and $(p \vee q)$. Also, remember to carefully check each and every condition of the two truth tables sequentially to prove them equivalent to each other.
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