I Translate compound proposition p → q (implication) to p↓q question

[VinnyW]

TL;DR 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 )

 

[Office_Shredder]

I'm confused, the last expression has no dependency on p at all? That can't possibly be the same thing logically.

 

[Mark44]

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.

p ↓ p ≡ ¬ p

What does the notation p ↓ p mean, particularly the down arrow? I've never seen that notation before.

 

[Office_Shredder]

It means nor

 

[SSequence]

The following page should be useful in that case:
https://en.wikipedia.org/wiki/NOR_logic

In "boolean algebra" notation we get the following as equivalent (following the above page):
$p \rightarrow q$
$p'+q$
$(p \,\, \mathrm{nor} \,\, p)+q$

Now one can expand the last expression using the "OR gate" equivalence.