translate compound proposition p → q to p ↓ q question (discrete math, logic)

VinnyW

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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 )
 
Make a truth table and check.
(or if you wish it directly):

[MATH]p \rightarrow q\\ \lnot p \lor q\\ ((T \land \lnot p) \lor q) \land T \\ (\lnot (F \lor p) \lor q) \land T \\ \lnot(\lnot(\lnot(F \lor p) \lor q) \lor F)\\ F \downarrow ((F\downarrow p) \downarrow q)[/MATH]
 
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