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To prove equivalence relation

  • Thread starter soopo
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  • #1
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Homework Statement



We have an equivalence relation such that
A <-> B.

Prove that the equivalence relation is true.

The Attempt at a Solution



Let
P: A -> B
Q: B -> A

Let's prove the relation by contradiction.
Assume
[tex]\neg A -> \neg B [/tex]

The previous assumption is the same as Q. Thus, we have a contradiction, since
it is impossible that both of the following Q and [tex] \neg Q[/tex]
are true at the same time, where

Q: B -> A and
[tex]\neg Q: \neg A -> \neg B [/tex] which is the same as B -> A.

Thus, the equivalence relation is true between A and B.
 

Answers and Replies

  • #2
290
2

Homework Statement



We have an equivalence relation such that
A <-> B.

Prove that the equivalence relation is true.
Is A <-> B given or is that what you're trying to prove?
 
  • #3
492
0
Are you trying to show that <-> is an equivalence relation on propositions?

If that is the case, you have to show the following three things:

A <-> A for all A.
if A <-> B then B <-> A.
if A <-> B AND B <-> C then A <-> C.

How should you show these? I'd probably use truth tables and the definition if <-> in terms of -> and "AND".
 

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