Total relation

**gordo151091** · April 2nd 2011, 12:35 PM

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?

**emakarov** · April 2nd 2011, 12:40 PM

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 $L\circ R$ ?

**gordo151091** · April 2nd 2011, 01:17 PM

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

$R \equiv (\forall x |<img src=$ \exists z|: xRz))" alt="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.

**Plato** · April 3rd 2011, 03:49 AM

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 $(a,b)\in L\circ R.$

**emakarov** · April 3rd 2011, 11:34 AM

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 $(a,b)\in L\circ R$ .

**Plato** · April 3rd 2011, 12:05 PM

Thanks, that is correct. It was either to early or the cut and paste.

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