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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.2012-07-06
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    Could you provide some context?2012-07-06
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    It is related to the exercise II.6.7 of the Bourbaki's book .2012-07-06
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    Here is Google Books [link](http://books.google.com/books?id=IL-SI67hjI4C&pg=PA128) to the page with the exercise.2012-07-06
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    From Bourbaki's Hints of this exercise, I conclude the proposition in this questions.2012-07-06
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    But I can't prove it. And can't find any counter-example.2012-07-06
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    @Sencodian Sorry, my proof was wrong, and I don't have time to fix it now.2012-07-08
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    @dtldarek I have found a counter-example which shows that this proposition and the Bourbaki'exercise are false. See my answer of this question.2012-07-08

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