Equality of Towers in Set

Theorem
Let $X$ be a non-empty set.

Let $\struct {T_1, \preccurlyeq_1}$ and $\struct {T_2, \preccurlyeq_2}$ be towers in $X$.

Then either:


 * $\struct {T_1, \preccurlyeq_1} = \struct {T_2, \preccurlyeq_2}$

or:


 * $\struct {T_1, \preccurlyeq_1}$ is an initial segment of $\struct {T_2, \preccurlyeq_2}$

or:


 * $\struct {T_2, \preccurlyeq_2}$ is an initial segment of $\struct {T_1, \preccurlyeq_1}$

Proof
By the definition of tower:
 * $\struct {T_1, \preccurlyeq_1}$ and $\struct {T_2, \preccurlyeq_2}$ are well-ordered sets.

By Wosets are Isomorphic to Each Other or Initial Segments, either:
 * $(1): \quad$ the towers are order isomorphic to each other

or:
 * $(2): \quad$ one is order isomorphic to an initial segment in the other.

, in case $(2)$, assume that $\struct {T_1, \preccurlyeq_1}$ is order isomorphic to an initial segment in $\struct {T_2, \preccurlyeq_2}$.

By Order Isomorphism iff Strictly Increasing Surjection, there exists a strictly increasing mapping:


 * $i: \struct {T_1, \preccurlyeq_1} \to \struct {T_2, \preccurlyeq_2}$

such that $i \sqbrk {T_1}$, the image set of $T_1$ under $i$, is equal to $T_2$ or to an initial segment of $T_2$.

From Characterization of Strictly Increasing Mapping on Woset, we can characterize $i$ as follows:
 * $\forall t \in T_1: \map i t = \map \min {T_2 \setminus i \sqbrk {S_t} }$

and:
 * $i \sqbrk {S_t} = S_{\map i t}$

Define:


 * $Y = \set {y \in T_1: \map i y = y}$

Then for any $t \in T_1$ and $S_t \subseteq Y$:

So $\map i y = y$ for any initial segment $S_t \subseteq Y$.

By Induction on Well-Ordered Set, $Y = T$.

Thus $i$ is the identity mapping from $T_1$ onto its image.

Recall that the image set of $i$ is $T_2$ or an initial segment of $T_2$.

So $T_1 = T_2$ or an initial segment of $T_2$.