Let $$ G=\langle x_0,x_1,\dots\mid x_jx_i=x_ix_{j+1}\text{ for }i<j\rangle, $$ $$ H=\langle a,b\mid [ab^{-1},a^{-1}ba],[ab^{-1},a^{-2}ba^2]\rangle, $$ where $[x,y]$ is commutator.
These are both presentations of Thompson group $F$ and I want to show that they are indeed isomorphic. I can prove that $\phi:H\to G$ when $\phi(a)=x_0$ and $\phi(b)=x_1$ is homomorphism. I define inverse as
$$\psi(x_0)=a,\psi(x_1)=b,\psi(x_n)=a^{1-n}ba^{n-1}.$$
And here I have problem.
It is easy to show that it works for $i=0$. Having that we can always assume that $i=1$, so all we need is to check that homomorphism works for all $j$.
It is easy to check that it works for $j=2,3$ because $$\psi(x_1^{-1})\psi(x_j)\psi(x_1)(\psi(x_{j+1}))^{-1}$$ is one of the relators of $H$.
But I do not know how to work out induction for any larger $j$.
I would be thankful for help or recommending some source where it is proved.