Summary: How to simplify ((p ↓ p) ↓ q) ↓ (p ↓ p) ↓ q) ) to F ↓ (( F ↓ q ) ↓ q ), whereas p and q are atomic propositions and F, probably, is contradiction.
I hope someone can help me or point me in the right direction.
I am reading Discrete Mathematics with its Applications by Rosen. I am trying to self learn discrete math. I am actually able to do most questions but I have a question about a solution (not the question itself.)
The question is (Section 1.3 Foundations: Logic and Proofs. Question 51)
Question: Find a compound proposition logically equivalent to p → q using only the logical operator ↓
My answer:
I know
p → q ≡ ¬ p ∨ q
and
p ↓ p ≡ ¬ p
By combining them, I got the answer:
((p ↓ p) ↓ q) ↓ (p ↓ p) ↓ q) )
which is the same answer as the one in the solution manual; however, the manual also lists:
F ↓ (( F ↓ q ) ↓ q )
I know F is contradiction.
How can I simplify
((p ↓ p) ↓ q) ↓ (p ↓ p) ↓ q) )
to
F ↓ (( F ↓ q ) ↓ q )
I hope someone can help me or point me in the right direction.
I am reading Discrete Mathematics with its Applications by Rosen. I am trying to self learn discrete math. I am actually able to do most questions but I have a question about a solution (not the question itself.)
The question is (Section 1.3 Foundations: Logic and Proofs. Question 51)
Question: Find a compound proposition logically equivalent to p → q using only the logical operator ↓
My answer:
I know
p → q ≡ ¬ p ∨ q
and
p ↓ p ≡ ¬ p
By combining them, I got the answer:
((p ↓ p) ↓ q) ↓ (p ↓ p) ↓ q) )
which is the same answer as the one in the solution manual; however, the manual also lists:
F ↓ (( F ↓ q ) ↓ q )
I know F is contradiction.
How can I simplify
((p ↓ p) ↓ q) ↓ (p ↓ p) ↓ q) )
to
F ↓ (( F ↓ q ) ↓ q )