I'm just learning transfinite induction over the ordinals, and I'm finding it a bit difficult to organize proofs. There seems to be several ways to organize almost any proof, but only one of which leads some place. Is the following proof right? Can it be proved "better"?
Proposition. Let $\alpha, \beta, \gamma$ be ordinals with $\beta < \gamma$. Then $\alpha + \beta < \alpha + \gamma$.
Proof I. By induction over $\gamma$.
Assume $\beta < \delta \Rightarrow \alpha + \beta < \alpha + \delta$ for $\delta < \gamma$.
Show $\beta < \gamma \Rightarrow \alpha + \beta < \alpha + \gamma$
In this post I will only deal with the case where $\gamma$ is a successor ordinal, $\gamma = \delta + 1$. Assume $\beta < \delta < \delta + 1 =\gamma$. Then $\beta < \delta < \gamma$ implies $\alpha + \beta < \alpha + \delta < \alpha + \delta + 1$, or $\alpha + \beta < \alpha + \gamma$.
Proof II. Induction over $\alpha$.
- Assume $\beta < \gamma$
Assume $\delta + \beta < \delta + \gamma$ for $\delta < \alpha$.
Show $\alpha + \beta < \alpha + \gamma$.
Now I let $\alpha = \delta + 1$ (again, I'll only dealing with successor ordinals), and I need to show $(\delta + 1) + \beta < (\delta + 1) + \gamma$. This is where I get stuck using this organization.
Is there a "best" or more elegant way to prove this simple proposition on ordinal addition?