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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?
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2 Answers

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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.

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4 Comments

Can this be done without having the truth table?
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@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.

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Hi

in the edit there is a small typo

P → ~q is F but you wrote T

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

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

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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
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@Genius Edited , thanks for pointing out.
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Answer:

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