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

In summary: So we have:##((p'+p)' \,\, \mathrm{or} \,\, (p'+p)') \,\, \mathrm{or} \,\, q##Using De Morgan's laws we can simplify this to:##(p'+p) \,\, \mathrm{or} \,\, q##Which is equivalent to:##F \,\, \mathrm{or} \,\, q##And since ##F \,\, \mathrm{nor} \,\, q## is equivalent to ##q## we finally get:##q##.In summary, the conversation discusses finding a compound proposition logically equivalent to p → q using only the logical operator ↓. The solution manual provides the
  • #1
VinnyW
4
0
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 )
 
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  • #2
I'm confused, the last expression has no dependency on p at all? That can't possibly be the same thing logically.
 
  • #3
VinnyW said:
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.
 
  • #4
It means nor
 
  • #5
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.
 

1. What does "p → q" mean in logic?

In logic, "p → q" is read as "if p, then q" or "p implies q". This means that if p is true, then q must also be true. If p is false, then the truth value of q does not matter.

2. How is the implication symbol "→" different from other logical operators?

The implication symbol "→" is different from other logical operators in that it does not require both p and q to be true for the entire statement to be true. As long as p is true, the statement is considered true regardless of the truth value of q.

3. What is the meaning of the "↓" symbol in "p↓q"?

The symbol "↓" is the logical operator for "or" in logic. In the context of "p↓q", it means that either p or q (or both) must be true for the entire statement to be true.

4. Can you give an example of how to translate "p → q" to "p↓q"?

One example is "If it is raining, then the ground is wet" (p → q). This can be translated to "It is raining or the ground is wet" (p↓q).

5. How is "p → q" related to conditional statements in everyday language?

"p → q" is related to conditional statements in everyday language because it represents a cause-and-effect relationship. If p happens, then q will also happen. This is similar to how we use conditional statements in everyday language, such as "If I study, then I will pass the exam".

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