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Question

# Let p, q , r 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
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B
P and R only
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C
P and S only
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D
P,Q,R and S
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Solution

## The correct option is C P and S only P:[(⇁p∨q)∧(r→s)∧(p∨r)]→(⇁s→q) ≡[(p→q)∧(r→s)∧(p∨r)]→(q∨s) which is a rule of inference called constructive dilemma and therefore valid. S:[p∧(p→r)∧(q∨⇁r)]→q ≡p(p′+r)(q+r′)→q ≡pr(q+r′)→q ≡prq→q ≡(prq)′+q ≡p′+r′+q′+q ≡p′+r′+1 ≡1 Therefore S is valid . Q and R can be similary simplified in boolean algebra to show that they are both not equivalent to 1 . So only P and S are valid. ​​​​​​​

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