Let $\alpha, \beta$ be ordinals.
$$\alpha + \beta:= \operatorname{Ord}(A, \leq)$$
where $A= (\alpha \times \{0\}) \cup (\beta \times \{1\})$ and where $\leq$ is the lexicographic order on $A$.
I want to show (with this definition only!)
$$(\alpha+ \beta)+ \gamma = \alpha + (\beta + \gamma)$$
Maybe I can show that the left side and the right side are isomorphic well-orders? I can see why this intuitively holds: both sides are just a string with $\alpha, \beta, \gamma$ after each other, but I can't make this formal.
Best Answer
The left hand side expands to $$(((\alpha\times\{0\})\cup(\beta\times\{1\}))\times\{0\})\cup(\gamma\times\{1\}),$$ while the right hand side expands to $$(\alpha\times\{0\})\cup(((\beta\times\{0\})\cup(\gamma\times\{1\}))\times\{1\})$$ and an isomorphism from the former to the latter is given by the function $f$ defined by $f(a,0,0)=(a,0)$ for $a\in\alpha$, $f(b,1,0)=(b,0,1)$, for $b\in\beta$, and $f(g,1)=(g,1,1)$ for $g\in\gamma$.
Of course this is just all a messy way to say that in the left hand side you're putting a copy of $\gamma$ after a copy of "a copy of $\beta$ after a copy of $\alpha$", while on the right hand side you're putting a copy of "a copy of $\gamma$ after a copy of $\beta$" after a copy of $\alpha$ and there should obviously be no difference between the two.