How to Prove the following Function Properties Definition: F is a function iff F is a relation and $(x,y) \in F$ and $(x,z) \in F \implies y=z$.  
I'm reading Introduction to Set Theory by Monk, J. Donald (James Donald), 1930 and i came across a theorem 4.10.  
Theorem 4.10  
(ii)$0:0 \to A$, if $F : 0 \to A$, then $F=0$.  
(iii) If $F:A\to 0$, then $A=F=0$.  
Where the book just explain the concept of function and now is stating its function property. I am stuck on what actually does it mean and how to prove it. May be can give me a hint.  
Thanks ahead.  

Consider the function $F\colon 0\to A$, suppose there is some $\langle x,y\rangle\in F$. This means that $x\in dom F$, since we have $dom F = 0$ then $x\in 0$ which is a contradiction. Therefore there are no ordered pairs in $F$, from the fact that it is a function we know that there are not other elements in $F$.
If so, we proved $F=0$.
The same proof goes for the other statement.
Edit:
An alternative method is by cardinal arithmetic: $|F|=|dom F| \le |dom F|\times|rng F|$
The first equality is simply by projection $\langle x,F(x)\rangle \mapsto x$, where the second is by the identity map.
From this, suppose $dom F = 0$ then $F=0$ and suppose $rng F=0$ then $F=0$ and $dom F=0$.
