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Suppose that the statement $p \rightarrow \neg q$ is false. What is the number of all possible combinations of truth values of $r$ and $s$ for which $(\neg q \rightarrow r) \wedge (\neg p \vee s)$ is true?

## 2 Answers

Best answer

The proposition p→q is False if p is True and q is False. In the question, it is given that p→¬q is False. Let us first find for what values of p and q, the proposition p→¬q will result in False.

 p q ¬q p → ¬q T T F F T F T T F T F T F F T T

From the above truth table we can observe that for only one combination when p is T and q is T, then proposition p→¬q is F.

Now, we have to find the number of all possible combinations of truth values of r and s for which the proposition (¬q→r)∧(¬p∨s) is true. The proposition (¬q→r)∧(¬p∨s) will be true when both (¬q→r) is true and (¬p∨s) is true.

 p q r s ¬p ¬q (¬q→r) (¬p∨s) (¬q→r)∧(¬p∨s) T T T T F F T T T T T T F F F T F F T T F T F F T T T T T F F F F T F F

So we can clearly see that for two combinations of r and s the proposition (¬q→r)∧(¬p∨s) is True i.e. when r = T and s = T, and r = F and s = T.

Edit:

To solve this question in fastest way, we first find when p→¬q is F, this is possible when p is T and q is T (one can take all four possible combinations TT, TF, FT, FF and solve on the fly by substitution).

• Now, for the proposition (¬q→r)∧(¬p∨s) will be true when both (¬q→r) is true and (¬p∨s) is true as there is an “AND” operator in between them.
• Also, the 1st part of this proposition (¬q→r) will always be T since q is T and so ¬q is always F and F → (T or F) is always T. Hence for all values of r, the 1st part always results in T.
• Now looking at the 2nd part (¬p∨s), this will be T when s is T as p is always T and so ¬p will always be F. So F∨(T or F) will be T if 2nd operator of this expression is T.
• This boils down to only two possible combinations of r and s when the proposition  (¬q→r)∧(¬p∨s) is T, which are r = T, s = T and r = F and s = T.

You can solve this question without using truth tables as above once you have practiced enough questions.

by

### 4 Comments

Can this be done without having the truth table?

@Arjun, I have edited my answer and added explanation to solve this question without using truth table method. Please let me know if the answer needs any changes.

Hi

in the edit there is a small typo

P → ~q is F but you wrote T

BTW excellent answer 👌
Thanks a lot @[ Jiren ] for pointing out the typo !!

Simple answer

p → ¬q is false : thus p is True and q is True

(¬𝑞→𝑟)∧(¬𝑝∨𝑠)  Ξ  (F→r)∧(F∨𝑠)  Ξ  (T ∧ s)  Ξ  s

Thus , 2​​​​​​ combinations with s as True and r as True/False

### 2 Comments

there will be 2 combinations

s can take only true and r can take true/false

Total combination = s’s Total Combination * r’s Total combination = 1*2 = 2
@Genius Edited , thanks for pointing out.
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