In this post, we will determine whether the following argument is valid or invalid. Lets consider the following two examples.
Question 1: Check whether the argument is valid or not:
p∨q, q→r, p→s, ∼s ⇒ r.
Answer:
We will proceed as follows:
1. p∨q (given)
2. q∨p (commutativity)
3. q→r (given)
4. ∼q∨r (equivalent to 3)
5. p∨r (applying Resolution to 2 and 4)
6. p→s ≡ ∼p∨s (given)
7. r∨s (applying Resolution to 5 and 6)
8. ∼s (given)
Now applying Disjunctive Syllogism to 7 and 8, we deduce r.
Therefore, the argument p∨q, q→r, p→s, ∼s ⇒ r is valid.
Question 2: Check whether the argument is valid or not:
p∨q, p∨q→∼w, ∼w→u∧∼v, u∧∼v→r∨s ⇒ r∨s.
Answer:
We will proceed as follows:
1. p∨q (given)
2. p∨q→∼w (given)
3. ∼w (applying Modulus Ponens to 1 and 2)
4. ∼w→u∧∼v (given)
5. u∧∼v (Modulus Ponens to 3 and 4)
6. u∧∼v→r∨s (given)
7. r∨s (applying Modulus Ponens to 5 and 6)
Therefore, r∨s is true, that is, we have shown that the argument p∨q, p∨q→∼w, ∼w→u∧∼v, u∧∼v→r∨s ⇒ r∨s is valid.
This article is written by Dr. Tathagata Mandal, Ph.D in Mathematics from IISER Pune (Algebraic Number Theory), Postdocs at IIT Kanpur & ISI Kolkata. Currently, working as an Assistant Prof. at Adamas University. Thank you for visiting the website.