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Let $p, q, r$ and $s$ be four primitive statements. Consider the following arguments:

  • $P:  [(¬p\vee  q) ∧ (r → s) ∧ (p \vee  r)] → (¬s → q)$
  • $Q:  [(¬p ∧q) ∧ [q → (p → r)]] → ¬r$
  • $R:  [[(q ∧ r) → p] ∧ (¬q \vee  p)] → r$
  • $S:  [p ∧ (p → r) ∧ (q \vee  ¬ r)] → q$


Which of the above arguments are valid?

  1. $P$ and $Q$ only
  2. $P$ and $R$ only
  3. $P$ and $S$ only
  4. $P, Q, R$ and $S$
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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)

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