October 4, 2023
October 4, 2023
October 4, 2023
###### Programming
October 4, 2023
 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?

 A P and Q only B P and R only C P and S only D 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.
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.