The set $T=\{l\in\mathbb{N}: ml=nl \ \text{implies} \ m=n \}$ is inductive. I'm trying to prove the following statement:
$ml=nl$ implies $m=n$ for every $m,n,l\in \mathbb{N}$. 
So I defined the set  $T=\{l\in\mathbb{N}: ml=nl \  \text{implies} \ m=n \}$ and if I prove that $T$ is inductive, then I done. 
It is clear that $1\in T$. Suppose that $l\in T$, then I have that $ml=nl$ implies $m=n$.
Now, if $m(l+1)=n(l+1)$ or $ml+m=nl+n$ how can one show that it implies $m=n$? 
 A: Let define the set $T$ bit differently, $T=\{m\in \mathbb{N}: ml=nl \ \  \text{implies} \ \ m=n \}$. We shall prove that $T$ is inductive.
First we show that $1\in T$. Assume that $m=1$, and $1\cdot l=n\cdot l$. If $n=1$, then $m=n$, and thus $1\in T$. If $n\neq 1$, then $n$ is in the range of function $S:\mathbb{N}\to \mathbb{N}$ (that is the function that maps $n$ to $n+1)$, and hence $n\in S(\mathbb{N})$, thus exist a $q\in\mathbb{N}$, s.t. $n=S(q)$. Now, $1\cdot l =n\cdot l=S(q)\cdot l=(q+1)\cdot l=q\cdot l+1\cdot l$.
Now we will look at the flowing equality:
$$1+1\cdot l=1+q\cdot l+1\cdot l$$
by the cancellation law of addition, we have that $1=1+q\cdot l$, or(by commutative law) $1=q\cdot l+1$, in other words $1=S(q\cdot l)$, which cannot be possible, since $1$ is the only element that is not in the range of $S$. Therefore this contradiction proves that $n=1$
Assume that $m=w\in T$, that is, $w\cdot l=n\cdot l$ implies $w=n$. Assume that $a,b\in\mathbb{N}$ such that $S(w)\cdot b=a\cdot b$. We show that $S(w)=a$. If $a=1$, then since $1\in T$, $S(w)=1$, which is impossible as $S(w)$ is in the range of $S$. Therefore we must have $a\neq 1$, that implies $a=S(r)$ for some $r\in\mathbb{N}$. Therefore we have 
$$S(w)\cdot b=S(r)\cdot b$$
Thus, $w\cdot b+b=r\cdot b+b$, and again by the cancellation law, $w\cdot b=r\cdot b$. By induction hypothesis $w=r$ and so(by the fact that $S$ is injective) we have $S(w)=S(r)$, which is also our $a$. Thus, $m=w+1\in T$; and so $T=\mathbb{N}$.
