Proof of Cancellation Law for Multiplication in N (Induction) I'm struggling a bit to prove the Cancellation Law for multiplication for the Natural numbers, using induction.
So far I have this:

Having $a, b \in N$ e $S = \lbrace c \in N | a \cdot c = b \cdot c \longrightarrow a = b \rbrace$
We know that $1 \in S$, since $a \cdot 1 = a = b = b \cdot 1$, 0 is not the base case because even if $a \neq b$, $a \cdot 0 = 0 = b \cdot 0$.
Using $k \in S$, by definition $a \cdot k = b \cdot k \longrightarrow a=b$, we have to show that $k' \in S$
Assuming that $a \cdot k' = b \cdot k'$, we have $a \cdot k + a = b \cdot k + b$, and I don't know where to go from here.

I tried using something like this, however I'm still stuck.
Thanks for any help
 A: HINT
The key is Trichotomy! $\forall x \forall y (x < y \lor y <x \lor x=y)$
Also, your theorem should be that $\forall a \forall b \forall c (c \not = 0 \rightarrow (a \cdot c = b \cdot c \rightarrow a = b))$
That is, for any $c \not = 0$:
If $a < b$ you can show that $a \cdot c < b \cdot c$, and for $b < a$ you have $b \cdot c < a \cdot c$, so if $a\cdot c = b \cdot c$ you must have $a=b$
A: There is another approach which requires defining a map:-
Let $x,y\in\mathbb{N}$. Define set $T=\{z\in\mathbb{N}\setminus\{0\}:xz=yz \implies x=y\}$.
Instead of showing that $T=\mathbb{N}\setminus\{0\}$, we define $S=T\cup\{0\}$ and show that $S=N$ by peano's inductive axiom.
Let $s:\mathbb{N}\to \mathbb{N}$ be the successor function with properties as defined in Peano's $5$ axioms characterizing $\mathbb{N}$.
$0\in S$ (by construction).
Assume for $z\in\mathbb{N}$, we have $z\in S$, i.e., $xz=yz \implies x=y$.
Define map $f:\mathbb{N}\to\mathbb{N}$ as $f(n)=n\times s(z)$, where $z$ is a fixed number as assumed above.
It is easy to show that $f$ is injective.
(Inductive step): $x\cdot s(z) = y\cdot s(z)$
$\implies f(x)=f(y)$
$\implies x=y$ ($\because f$ is injective).
Thus, we have shown: $z\in S \implies s(z)\in S$. Hence, $S=\mathbb{N}$.
