$\alpha < \beta \implies \gamma+\alpha < \gamma + \beta$ for ordinals 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.
 A: Let $f: (\gamma \times \{0\}) \sqcup (\alpha \times \{1\}) \to (\gamma \times \{0\}) \sqcup (\beta \times \{1\})$ be an order isomorphism. We show by induction on $\delta < \gamma$ that $f(\delta,0) = (\delta,0)$. 
Base case: $f$ is an order isomorphism, so it must send the least element of $(\gamma \times \{0\}) \sqcup (\alpha \times \{1\})$ to the least element of $(\gamma \times \{0\}) \sqcup (\beta \times \{1\})$. The least element in either case is $(0,0)$. 
Induction step: Suppose $f$ is the identity below $(\delta,0)$. Because $f$ is an order isomorphism, it must map $(\delta,0)$ to the least element of $(\gamma \times \{0\}) \sqcup (\beta \times \{1\})$ not in $f[\delta \times \{0\}]$; but $f[\delta \times \{0\}] = \delta \times \{0\}$ by hypothesis, so $f(\delta,0) = (\delta,0)$.
Thus, by induction $f$ is the identity on $\gamma \times \{0\}$. It follows that $f|_{\alpha \times\{1\}}$ must be an isomorphism of $\alpha \times \{1\}$ onto $\beta \times \{1\}$ for $f$ to be an isomorphism (injectivity implies elements of $\alpha \times \{1\}$ must map into $\beta \times \{1\}$, surjectivity implies all element of $\beta \times \{1\}$ are mapped to in this way, and $f$ being order-preserving implies that $f|_{\alpha \times \{1\}}$ is as well). Contradiction.
