Sorry for the block of text from Kirill Mackenzie's chapter on groupoids.
I have proven that $\natural,\natural_0$ is a groupoid morphism and that this quotient construction satisfies the groupoid axioms. I am having trouble showing that the kernel is $\Phi$.
Suppose $\xi\in \ker(\natural)$. Then $\exists x\in B$ such that $[x]=\natural_0(x)\in B/\Phi$ and $\natural(\xi)=[\xi]=[\tilde{x}]$. Now $[\xi]=[\tilde{x}]\Leftrightarrow \zeta,\zeta'\in \Phi$ such that $\xi=\zeta\tilde{x}\zeta'=\zeta\zeta'$, where the last equality comes from the fact that if $\zeta\tilde{x}\zeta'$ is defined then $\alpha(\zeta)=\beta(\tilde{x})=x$ so $\zeta\tilde{x}=\zeta\widetilde{\alpha(\zeta)}=\zeta$. Now $\xi=\zeta\zeta'\in \Phi$ since $\Phi$ is a subgroupoid. I think this shows $\ker(\natural)\subseteq \Phi$. I am having a trouble showing the reverse inclusion.
The only insights I can see that might help are:
Let $\xi\in \Phi$ with $\alpha(\xi)=x$,$\beta(\xi)=y$. $\xi = \xi\xi^{-1}\xi$, so $[\xi]=[\xi^{-1}]$ and hence $\natural(\xi)=\natural(\xi^{-1})$.
$\bar{\alpha}(\natural(\xi))=\natural_0(\alpha(\xi))=[x]$
$\bar{\alpha}(\natural(\xi^{-1}))=\natural_0(\alpha(\xi^{-1}))=[y]$
Hence $[x]=[y]$ and so $x\sim y$ in B. This seems like it might be useful but I am completely stuck here and not sure how to show that $\natural(\xi)=[\tilde{x}]$ for some $x\in B$.
I think I got it but it's very unsatisfying and kind of fishy:
Let $\xi\in \Phi$. Then $\widetilde{\alpha(\xi)} = \xi^{-1}\xi\widetilde{\alpha(\xi)}$, and since $\xi^{-1},\widetilde{\alpha(\xi)}\in \Phi$, so $[\widetilde{\alpha(\xi)}]=[\xi]=\natural(\xi)$. Hence $\xi\in \ker(\natural)$.
$\therefore \Phi=\ker(\natural)$.