Let’s try to make conclusion false and all the premises true of the argument. If it’s not possible then it is a valid argument, otherwise invalid argument.
P : [(¬p v q) ∧ (r → s) ∧ (p v r)] → (¬s → q)
Conclusion is false when s=F and q=F. Now let’s try to make each premise true. (¬p v q) is true when p=F. (r → s) is true when r=F. But now we can’t make (p v r) = true, because p=F & r=F => (p v r) = F. So, the argument P is valid. (not possible to make conclusion false and all the premises true of the argument)
Q : [(¬p ∧ q) ∧ (q → (p → r))] → (¬r)
Conclusion is false when r=T. Now let’s try to make each premise true. (¬p ∧ q) is true when p=F and q=T. (q → (p → r)) = (q → (p → T)) = (q → T) is true. So, the argument Q is invalid. (possible to make conclusion false and all the premises true of the argument)
R: [[(q ∧ r) → p] ∧ (¬q v p)] → r
Conclusion is false when r=F. Now let’s try to make each premise true. [(q ∧ r) → p] = [F → p] is true. (¬q v p) is true for 3 possible combinations of p & q. So, the argument Q is invalid. (possible to make conclusion false and all the premises true of the argument)
S: [p ∧ (p → r) ∧ (q v ¬ r)] → q
Conclusion is false when q=F. Now let’s try to make each premise true. (q v ¬ r) is true when r=F. (p) is true when p=T. But now we can’t make (p → r) = true, because p=T & r=F => (p → r) = F. So, the argument P is valid. (not possible to make conclusion false and all the premises true of the argument)
