Part 2 Proof By Contradiction

Question 3

Let us prove that

(p q) (pq)

is a logical law, i.e. it is always true.

We will assume (p q) (pq) can be false, and show that this assumption leads to a Contradiction.

Please insert truth values (1 or 0 after =) below.

1. Assume (p q) (pq)=0
2. p q= [1]
3. (pq)= [1]
4. pq= [3]
5. p = [2]
6. q= [2]
7. p= [4]
8. q= [5]
9. pq= [6,7]

Do you see a Contradiction?

Please fill in 1-9 with 0 or 1