If $\alpha \preceq \beta$ then there is a unique ordinal $\gamma$ such that $\alpha + \gamma = \beta$ I tried to prove this proposition, if $\alpha \preceq \beta$ then there is a unique ordinal $\gamma$ such that $\alpha + \gamma = \beta$, by order type not by transfinite induction.
My attempt:

*

*$W = \{ \xi \in \beta \mid \alpha \preceq \xi \}$ is a set of ordinals so, $W$ is a woset.

*Every woset is order isomorphic to a unique ordinal, therefore, exist a unique ordinal $\gamma$ such that, $W$ and $\gamma$ are ordered isomorphic.

My problem:
I strongly believe that, $\gamma$ is the solution but I cannot prove that, $\alpha + \gamma = \beta$.
Def: $\prec := \in$. The ordinal addition is defined by induction.
I have read some similar questions. For example, (1) just talk about transfinite induction (limit case) and (2) just talk about the uniqueness. I hope your helps.
 A: Your demonstration that $W$ has the order type of an ordinal is correct. Once you have that $W$ is isomorphic to a unique ordinal $\gamma$, you're basically done, as $\alpha + \gamma = \beta$ by the definition of ordinal addition:

The union of two disjoint well-ordered sets $S$ and $T$ can be well-ordered. The order-type of that union is the ordinal which results from adding the order-types of $S$ and $T$. If two well-ordered sets are not already disjoint, then they can be replaced by order-isomorphic disjoint sets, e.g. replace $S$ by $\{ 0 \} × S$ and $T$ by $\{ 1 \} × T$. This way, the well-ordered set $S$ is written "to the left" of the well-ordered set T, meaning one defines an order on $S \cup T$ in which every element of $S$ is smaller than every element of $T$. The sets $S$ and $T$ themselves keep the ordering they already have.

By your definition of $W$, $$W := \{ \xi \in \beta | \alpha = \xi \text{ or } \alpha \in \xi \},$$ it's clear that $\beta$ is the disjoint union of $\alpha = \{ \xi \in \beta | \xi \in \alpha \}$ with $W$.
But the disjoint union of $\alpha$ with $W$ is order-isomorphic to the disjoint union of $\{ 0 \} \times \alpha$ with $\{ 1 \} \times \gamma$ equipped with the dictionary order, since $\gamma$ is the unique ordinal with the order type of $W$. This is the definition of what it means for $\alpha + \gamma$ to equal $\beta$.
If you do it this way, no induction argument is required. The heart of the question is whether there is an argument that works--and I quote--

"by order type not transfinite induction,"

and this would seem to be what the original post is asking for.
A: I have proved one theorem:
If two disjoint ordered sets, $(W_1, \preceq_1)$ order isomorphic to $\alpha$ and $(W_2, \preceq_2)$ order isomorphic to $\gamma$, then ordered sum of the two ordered sets, $(W_1, \preceq_1) \oplus (W_2, \preceq_2)$, order isomorphic to $\alpha + \gamma$.
Def: $\forall x,y \in (W_1 \cup W_2, \preceq) = (W_1, \preceq_1) \oplus (W_2, \preceq_2): x \preceq y$ is defined:

*

*$x,y \in W_1 \wedge x \preceq_1 y$ or

*$x,y \in W_2 \wedge x \preceq_2 y$ or

*$x \in W_1 \wedge y \in W_2$.

Back to the problem above, from $\alpha \cup W = \beta$ and $W$ order isomorphic to $\gamma$ so, $\alpha \oplus W = \beta$ order isomorphic to $\alpha + \gamma$ by the theorem. Therefore, $\alpha + \gamma = \beta$.
Just find out a way to explain.
