
If the truth value of the statement $p\to \left( \sim q\vee r \right)$ is false (F), then the truth values of the statements p, q, r are respectively
(A) F, T, T
(B) T, F, F
(C) T, T, F
(D) T, F, T
Answer
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Hint: We start solving this problem by making the truth value tables for $\sim q\vee r$ and then making the truth value table for $p\to \left( \sim q\vee r \right)$. Then we look at that truth value table for the truth values of p and $\sim q\vee r$. Then after finding the truth value of $\sim q\vee r$, we use the truth value table of $\sim q\vee r$ to find the truth values of q and r.
Complete step by step answer:
We are asked to find the truth values of p, q, r when the truth value of $p\to \left( \sim q\vee r \right)$ is false.
Let us go through the truth table of $a\vee b$.
Using that we can write the truth table of $\sim q\vee r$
Now, let us go through the truth table of $a\to b$
Using this we can write the truth value table for $p\to \left( \sim q\vee r \right)$
As we are given that the truth value of $p\to \left( \sim q\vee r \right)$ is False. From the table we can say that it is possible only when the truth value of p is True and truth value of $\sim q\vee r$ is False.
From the truth table of $\sim q\vee r$ we can see that it is false only when the truth values of q and r are true and false respectively.
So, we get that truth value of $p\to \left( \sim q\vee r \right)$ is False when the truth values of p, q, r are True, True and False respectively.
So, the correct answer is “Option C”.
Note: The mistake that one makes while solving this problem are one might take the truth value table for $a\vee b$ as
But it is the truth value table for $a\wedge b$. One must note the difference between $a\vee b$ and $a\wedge b$. The symbol $\wedge $ represents AND while the symbol $\vee $represents OR.
Complete step by step answer:
We are asked to find the truth values of p, q, r when the truth value of $p\to \left( \sim q\vee r \right)$ is false.
Let us go through the truth table of $a\vee b$.
| a | b | $a\vee b$ |
| F | F | F |
| F | T | T |
| T | F | T |
| T | T | T |
Using that we can write the truth table of $\sim q\vee r$
| q | $\sim q$ | r | $\sim q\vee r$ |
| F | T | F | T |
| F | T | T | T |
| T | F | F | F |
| T | F | T | T |
Now, let us go through the truth table of $a\to b$
| a | b | $a\to b$ |
| F | F | T |
| F | T | T |
| T | F | F |
| T | T | T |
Using this we can write the truth value table for $p\to \left( \sim q\vee r \right)$
| p | $\sim q\vee r$ | $p\to \left( \sim q\vee r \right)$ |
| F | F | T |
| F | T | T |
| T | F | F |
| T | T | T |
As we are given that the truth value of $p\to \left( \sim q\vee r \right)$ is False. From the table we can say that it is possible only when the truth value of p is True and truth value of $\sim q\vee r$ is False.
From the truth table of $\sim q\vee r$ we can see that it is false only when the truth values of q and r are true and false respectively.
So, we get that truth value of $p\to \left( \sim q\vee r \right)$ is False when the truth values of p, q, r are True, True and False respectively.
So, the correct answer is “Option C”.
Note: The mistake that one makes while solving this problem are one might take the truth value table for $a\vee b$ as
| a | b | $a\vee b$ |
| F | F | F |
| F | T | F |
| T | F | F |
| T | T | T |
But it is the truth value table for $a\wedge b$. One must note the difference between $a\vee b$ and $a\wedge b$. The symbol $\wedge $ represents AND while the symbol $\vee $represents OR.
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