I'm trying to prove if $\beta < \gamma$, then $\alpha + \beta < \alpha + \gamma$ for ordinals $\alpha, \beta, \gamma$ via induction on $\gamma$.
First, if $\gamma = 0$, the statements vacuously holds because there are no such $\beta < \gamma$.
I run into a problem for the case $\gamma = \gamma' + 1$, the successor case. The IH would be that for $\beta' < \gamma'$, $\alpha + \beta' < \alpha + r'$. Now, I want to show for $\beta < \gamma, \alpha + \beta < \alpha + \gamma$. How do I quantify over $\beta$ in this case?
Any advice would be appreciated!
This post Prove that if $\alpha<\beta$ then $\gamma + \alpha < \gamma + \beta$ for ordinals. asks the same question, but it doesn't address the successor case.