# Total relation

• Apr 2nd 2011, 12:35 PM
gordo151091
Total relation
How can I proof that if $L$ is a universal relation that $R total relation\equiv L = L \circ R$. It is pretty obvious that it is true, but how do I prove it?
• Apr 2nd 2011, 12:40 PM
emakarov
Could you explain what Rtotalrelation is? Also, is universal relation the one that contains all possible pairs? Finally, by $L\cdot R$ do you mean the composition $L\circ R$?
• Apr 2nd 2011, 01:17 PM
gordo151091
Quote:

Originally Posted by emakarov
Could you explain what Rtotalrelation is? Also, is universal relation the one that contains all possible pairs? Finally, by $L\cdot R$ do you mean the composition $L\circ R$?

R total relation is

$R \equiv (\forall x |:(\exists z|: xRz))$

Yes, the universal relation is the one that contains all pairs

Yes, I did mean the composition. Edited the OP.
• Apr 3rd 2011, 03:49 AM
Plato
This is just a matter of using the notation.
If $(a,b)$ is any pair then $\left( {\exists c} \right)\left[ {(c,b) \in R} \right]$ because $R$ is total.
Because $L$ is universal $(a,c)\in L$ so $(a,b)\in L\circ R.$
• Apr 3rd 2011, 11:34 AM
emakarov
Quote:

Originally Posted by Plato
If $(a,b)$ is any pair then $\left( {\exists c} \right)\left[ {(c,b) \in R} \right]$ because $R$ is total.
Because $L$ is universal $(a,c)\in L$ so $(a,b)\in L\circ R.$

I think it should say that for any (a,b) there exists a $c$ such that $(a,c)\in R$ because R is total. Then $(c,b)\in L$ because L is universal, so $(a,b)\in L\circ R$.
• Apr 3rd 2011, 12:05 PM
Plato
Thanks, that is correct. It was either to early or the cut and paste.