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.
