Does the existence of only right neutral mean that there can only exist right inverse? Let $(G,\circ $) be groupoid. If there exists only right neutral, does that mean that there can exist only right inverse?
To put it in another way, does the existence of only right neutral mean that there can not exist left inverse?
I assumed this from an example where $\circ$ is operation of exponentiation and is defined by: $$(\forall x,y\in G) x\circ y=x^y$$
So I want to make sure my assumption is correct. 
Edit: Definition of groupoid I'm using:  

Ordered pair $(G,\circ)$ where $G$ is a set, and $\circ$ is an internal binary operation is called groupoid.

 A: After further research I found out that existance of only right neutral doesn't mean there can't exist left inverse. I found this out by observing  operation of division $/$ on a set of positive rational numbers $Q^+$.
The operation of division has only right neutral, but has both left and right inverse.  
Here is my proof: 
Let $(G,\circ)$ be groupoid (or magma as suggested) where $G=Q^+$ and $\circ =/$.
$$(\exists e\in G)(\forall x\in G) e\circ x=x\circ e=e$$
For left neutral:
$$e\circ x=x$$
$$e\circ x=\frac{e}{x} $$
from these two we get $x=\frac{e}{x}$ which means that there is no left neutral.
For right neutral:
$$x\circ e=x$$
$$x\circ e=\frac{x}{e}$$
from these two we get $x=\frac{x}{e}$ which means that there is only right neutral $e=1$.  
For inverse:
$$(\forall x\in G)(\exists x^{-1}\in G)x\circ x^{-1}=x^{-1}\circ x=e$$
For left inverse:
$$x^{-1}\circ x=e$$
$$x^{-1}\circ x=\frac{x^{-1}}{x}$$
from these two we get $\frac{x^{-1}}{x}=1\Rightarrow x^{-1}=x$  
For right inverse:
$$x\circ x^{-1}=e$$
$$x\circ x^{-1}=\frac{x}{x^{-1}}$$
from these two we get $\frac{x}{x^{-1}}=1\Rightarrow x^{-1}=x$.  
This proves there exist both left and right inverse $x^{-1}=x$ even though there exists only right neutral.
(At least I think this proves it, correct me if I'm wrong :) )
