I want to prove the following statement:



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> Let $u$ be a vertex in a $2$-connected graph $G$. Then $G$ has two spanning trees such that for every vertex $v$, the $u,v$-paths in the trees are independent.
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I tried to show this, but surprisingly, I have proved another statement.



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> A graph with $\vert V(G) \vert \geq 3$ is $2$-connected iff for any two vertices $u$ and $v$ in $G$, there exist at least two independent $u,v$-paths.
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And I can assure that it is true, since I could find it from other papers.  

I think this one may help me proving the desired statement, but I have no idea how to use it properly.  

Would you help me find a such way, or suggest another proof of the first statement?



Two independent spanning trees of $2$-connected graph
Yes, this is true. We will prove the following stronger lemma.


**Lemma.** Let $G$ be a $2$-connected graph and $u \in V(G)$. Then $G$ contains two spannings trees $T\_1$ and $T\_2$ such that for all $a,b \in V(G) \setminus \{u\}$ (possibly $a=b$), either


* the $ua$-path in $T\_1$ and the $ub$-path in $T\_2$ are internally-disjoint, or
* the $ub$-path in $T\_1$ and the $ua$-path in $T\_2$ are internally disjoint.


*Proof.* We proceed by induction on $|V(G)|+|E(G)|$. Since $G$ is $2$-connected, $G$ has an [ear decomposition](https://en.wikipedia.org/wiki/Ear_decomposition) $(C, P\_1, \dots, P\_k)$, with $u \in V(C)$. If $G=C$, let $e\_1$ and $e\_2$ be the two edges incident to $u$. It is easy to check that we may take $T\_1=C \setminus e\_1$ and $T\_2=C \setminus e\_2$. Thus, we may assume $k \geq 1$.
If the last ear $P\_k$ is just an edge $e$, then $G \setminus e$ is $2$-connected, and we can apply induction. Thus, we may assume $|V(P\_k)| \geq 3$. Suppose $P\_k=x\_1x\_2 \dots x\_\ell$. Observe that $G'=G-\{x\_2, \dots, x\_{\ell-1}\}$ is $2$-connected. By induction, $G'$ contains two spanning trees $T\_1'$ and $T\_2'$ such that for all $a,b \in V(G') \setminus \{u\}$, either


* the $ua$-path in $T\_1'$ and the $ub$-path in $T\_2'$ are internally-disjoint, or
* the $ub$-path in $T\_1'$ and the $ua$-path in $T\_2'$ are internally disjoint.


In particular, either


* the $ux\_1$-path in $T\_1'$ and the $ux\_\ell$-path in $T\_2'$ are internally-disjoint, or
* the $ux\_\ell$-path in $T\_1'$ and the $ux\_1$-path in $T\_2'$ are internally disjoint.


In the first case, $T\_1:=T\_1' \cup x\_1x\_2 \dots x\_{\ell-1}$ and $T\_2:=T\_2' \cup x\_2x\_3 \dots x\_\ell$ are the required spanning trees of $G$. In the second case, $T\_1:=T\_1' \cup x\_2x\_3 \dots x\_\ell$ and $T\_2:=T\_2' \cup x\_1x\_2 \dots x\_{\ell-1}$ are the required spanning trees of $G$.


