UGC NET CS 2014 June-paper-3
October 4, 2023Programming
October 4, 2023Engineering-Mathematics
| Question 5 |
Let p, q and s be four primitive statements. Consider the following arguments:
P: [(¬p ∨ q) ∧ (r → s) ∧ (p ∨ r)] → (¬s → q)
Q: [(¬p ∧ q) ∧ [q → (p → r)] → ¬r
R: [[(q ∧ r) → p] ∧ (¬q ∨ p)] → r
S: [p ∧ (p → r) ∧ (q ∨ ¬r)] → q
Which of the above arguments are valid?
| P and Q only | |
| P and R only | |
| P and S only | |
| P, Q, R and S |
Question 5 Explanation:
Let p→q is conditional proposition here. p and q are compound propositions itself. Arguments to be valid if all combinations have to be tautology (like T→T, F→T, F→F) and its invalid if it have fallacy (T→F).
If we somehow get this fallacy (T→F) then an argument is invalid.
For options P and S you don’t get any such combinations for T→F, so P and S are valid.
For option Q: If we put p=F, q=T, r=T then we get T→F. So its INVALID.
For option R: If we put p=F, q=F, r=F then we get T→F. So it is INVALID.
So, answer is (C).
If we somehow get this fallacy (T→F) then an argument is invalid.
For options P and S you don’t get any such combinations for T→F, so P and S are valid.
For option Q: If we put p=F, q=T, r=T then we get T→F. So its INVALID.
For option R: If we put p=F, q=F, r=F then we get T→F. So it is INVALID.
So, answer is (C).
Correct Answer: C
Question 5 Explanation:
Let p→q is conditional proposition here. p and q are compound propositions itself. Arguments to be valid if all combinations have to be tautology (like T→T, F→T, F→F) and its invalid if it have fallacy (T→F).
If we somehow get this fallacy (T→F) then an argument is invalid.
For options P and S you don’t get any such combinations for T→F, so P and S are valid.
For option Q: If we put p=F, q=T, r=T then we get T→F. So its INVALID.
For option R: If we put p=F, q=F, r=F then we get T→F. So it is INVALID.
So, answer is (C).
If we somehow get this fallacy (T→F) then an argument is invalid.
For options P and S you don’t get any such combinations for T→F, so P and S are valid.
For option Q: If we put p=F, q=T, r=T then we get T→F. So its INVALID.
For option R: If we put p=F, q=F, r=F then we get T→F. So it is INVALID.
So, answer is (C).
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