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Let $G_0$, $G_1$, $H_0$, $H_1$ be four equivalence relations on a set $E$ such that $G_1\cap H_0=G_0\cap H_1$ and $G_1\circ H_0=G_0\circ H_1$. Let $x\in E$. Prove that for every $y\in G_1(x)$, there exists a $z\in G_1(x)\cap H_1(x)$, such that $(y,z)\in G_0$; and for every $y\in H_1(x)$, there exists a $z\in G_1(x)\cap H_1(x)$, such that $(y,z)\in H_0$.

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I've tried to edit your question using TeX for better readability. (And maybe some other users will improve it a little more.) You should check whether I did not change the meaning of your question, unintentionally. If you're satisfied with the edited version, you can remove the original one. – Martin Sleziak Jul 6 '12 at 6:55
Could you provide some context? – dtldarek Jul 6 '12 at 7:15
It is related to the exercise II.6.7 of the Bourbaki's book <Theory of sets>. – Sencodian Jul 6 '12 at 8:02
Here is Google Books link to the page with the exercise. – Martin Sleziak Jul 6 '12 at 8:07
From Bourbaki's Hints of this exercise, I conclude the proposition in this questions. – Sencodian Jul 6 '12 at 8:11

I'm surprised that I find a counter-example which shows that this proposition and the Bourbaki'exercise are false! The following is my counter-example:

Let $E=\{1,2,3,4\}$, $x=2$,

$G_0=\{(1,1),(3,3),(1,3),(3,1),(2,2),(4,4),(2,4),(4,2)\}$, $G_1=\{(1,1),(2,2),(1,2),(2,1),(3,3),(4,4),(3,4),(4,3)\}$, $H_0=H_1=\{(1,1),(4,4),(1,4),(4,1),(2,2),(3,3),(2,3),(3,2)\}$.

Then we have:

$G_1\cap H_0=G_0\cap H_1=\{(1,1),(2,2),(3,3),(4,4)\}$,

$G_1\circ H_0=G_0\circ H_1=E\times E$.

But in this case the conclusion of this proposition and the Bourbaki's exercise are all false.

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