Re-write $1 \cdot x$ to $x$. Given the following bi-directional re-write rules (where $1$ is a constant, $^{-1}$ is a unary operator, $\cdot$ is a binary operator, and $x,y,z$ are arbitrary terms):
$$\begin{align*}
x \cdot 1 &= x \\
x \cdot (y \cdot z) &= (x\cdot y) \cdot z \\
x \cdot x^{-1} &= 1
\end{align*}$$
we're asked to prove that $1 \cdot x = x$ (ie, there is a chain $t_0 \to t_1 \to t_3 \to \cdots \to t_n$ with $t_0 = 1\cdot x$, $t_n = x$, and $t_i \to t_{i+1}$ meaning one of the 3 equations above re-writes $t_i$ to $t_{i+1}$ (in either direction)).
After staring at this for a while I'm beginning to doubt whether or not this is possible... can anyone a) confirm this is indeed possible and b) potentially nudge me in the right direction?

Another method of proof would be:
Using the fact that $x = (x^{-1})^{-1}$, then $x^{-1} \cdot x = 1$ from axiom 3.
\begin{align*}
&1 \cdot 1 = 1 \\
&1 \cdot (x \cdot x^{-1}) = 1 \\
&(1 \cdot x) \cdot x^{-1} = x \cdot x^{-1} \\
&((1 \cdot x) \cdot x^{-1}) \cdot x = (x \cdot x^{-1}) \cdot x \\
&(1 \cdot x) \cdot (x^{-1} \cdot x) = x \cdot (x^{-1} \cdot x) \\
&(1 \cdot x) \cdot 1 = x \cdot 1 \\
&1 \cdot x = x \\
\end{align*}
