Prove that for every three ordinals $\beta \lt \gamma \Rightarrow \alpha+\beta \lt \alpha+\gamma$ 
Prove that for every three ordinals $\alpha,\beta,\gamma$ we have $\beta \lt \gamma \Rightarrow \alpha+\beta \lt \alpha+\gamma$

It's obvious if all of them are finite, also if only alpha is finite then we have the left cancellation law. 
Now if all of them are infinite, this is what I tried: we know that $\displaystyle\sup_{\delta<\beta}(\delta)<\sup_{\xi<\gamma}(\gamma) \Rightarrow\alpha+\sup_{\delta<\beta}(\delta) \lt \alpha+\sup_{\xi<\gamma}(\xi)$ The alphas are equal so we get: $\displaystyle\sup_{\delta<\beta}(\delta)<\sup_{\xi<\gamma}(\xi)=\beta \lt \gamma$. 
Is there another way solve it without using the supermum ? Is my approach even correct ?

Pick representatives from $\alpha,\beta$ and $\gamma$ ($A,B$ and $C$). Let $f\colon B\to C$ be an embedding of $B$ into a proper initial segment of $C$.
Now try to find an embedding from $A+B$ into $A+C$ whose image is an initial segment.
Finally, recall that between two well-orders there is exactly one embedding mapping one into an initial segment of the other. Now use the fact that $f$ embeds $B$ into a proper initial segment of $C$, and finish the proof.
