Implications from propositions A => B, A <=> C and C => B

  • Thread starter Thread starter x-is-y
  • Start date Start date
AI Thread Summary
The discussion centers on the implications between three mathematical propositions: A (x^2 < 16), B (-4 < x), and C (-4 < x < 4). The key states that A implies B, A is equivalent to C, and C implies B, but there is confusion regarding whether these implications are sufficient to establish all possible relationships among the propositions. It is noted that proving A => B, B => C, and C => A would indeed demonstrate the equivalence of all three propositions, but the original key does not provide a complete set of implications. The participants express uncertainty about the phrasing of the question and the relevance of the initial statements to the task of listing all implications. Clarification is sought on how to interpret the requirement to "put out every possibility" regarding the implications.
x-is-y
Messages
5
Reaction score
0
I'm not 100% sure what this is in English so I'll try to describe it. Gives that:

A: x^2 < 16
B: -4 < x
C: -4 < x < 4

I'm supposed to put out every possibility for => and <=> between A,B and C. The key says that A => B, A <=> C and C => B. I can understand this, but isn't it true for every proposition (I think that it's called this) that A <=> A. That is, every proposition implies itself?
 
Physics news on Phys.org
Suppose you have three propositions, A, B and C and want to prove that they are equivalent. That means that
A <=> B, B <=> C and A <=> C
which can also be written as
A => B, B => A, A => C, C => A, B => C, C => B.
In words: if you know that one of them is true/false, they must all be true/false.
Obviously, A <=> A is always true and it's not included in the list.

You can prove all 6 of them consecutively, but that would be a lot of work. Therefore we find a shortcut:
suppose you would be able to prove half of them, namely that A => B, B => C and C => A.
Then because B implies C and C implies A, B also implies A so you automatically get B => A and therefore A <=> B.
Similarly, C => A and A => B so C also implies B (via A) hence B <=> C.
So proving these three statements will prove all six of them.

This is what the key suggests. It is wrong though, in claiming that A => B, A <=> C and C => B suffices. For example, you cannot get B => A (there is no premise starting with B).
 
I think you misunderstood the question. Nothing was said about A<=>B, B<=>C, C<=>A. there is nothing here that implies B=> A and the answer key does NOT suggest such a thing.
 
x-is-y said:
I'm supposed to put out every possibility for => and <=> between A,B and C. The key says that A => B, A <=> C and C => B.
Probably then I got confused by the formulation.
What is meant by "put out every possbility"?
And I don't think I see the relevance of the question in the first part of the post to the question "isn't it true for every proposition ...".

So I apologize if I mislead you x-is-y, perhaps you can try to rephrase the question (or if someone else understands it, explain it to my numb mind)?
 
Back
Top