Let $\alpha,\beta, \gamma$ be ordinals. I'm trying to prove that:
$$\alpha < \beta \implies \gamma+\alpha < \gamma + \beta$$
I managed to prove that
$$\alpha < \beta \implies \gamma + \alpha \leq \gamma + \beta$$
by showing there is a strictly order preserving function $\gamma+\alpha \to \gamma + \beta$
but I struggle to show equality is impossible.
So suppose to the contrary that:
$$\gamma+\alpha= \gamma + \beta$$
This implies that
$$(\gamma \times 0) \cup (\alpha \times 1) \cong (\gamma \times 0) \cup (\beta \times 1)$$
But I don't think we really can get a contradiction from this.
I read that ordinal addition is left addition, so we should somehow be able to prove that $\alpha = \beta$ which will be a good contradiction.