avewright/proofwiki-math huggingface dataset:
Theorem: Existence of Minimal Uncountable Well-Ordered Set

There exists a minimal uncountable well-ordered set.
That is, there exists an uncountable well-ordered set $\Omega$ with the property that every initial segment in $\Omega$ is countable.

Proof:
By the axiom of powers, there exists the power set $\powerset \N$, where $\N$ is the set of natural numbers.
By Power Set of Natural Numbers is not Countable, this set is uncountable.
By Zermelo's Well-Ordering Theorem, $\powerset \N$ can be endowed with a well-ordering.
Denote such an ordering with the symbol $\preccurlyeq$.
Let $\powerset \N_a$ denote the initial segments of $\powerset \N$ determined by $a \in \powerset \N$  
Suppose $\struct {\powerset \N, \preccurlyeq}$ has the property:
:''$\powerset \N_a$ is countable for every $a \in \powerset \N$''
Then set $\Omega = \powerset \N$.
Otherwise, suppose $\struct {\powerset \N, \preccurlyeq}$ does not have the above property.
Consider the subset of $\powerset \N$:
:$P \subseteq \set {a \in \powerset \N : \powerset \N_a \text{ is uncountable} }$
Then $P$ has a smallest element, by the definition of a well-ordered set.
Call such an element $a_0$.
That is, $a_0 \in \powerset \N$ is the smallest $a$ such that $\powerset \N_{a_0}$ is uncountable.
Then the initial segment $\powerset \N_{a_0}$ is itself uncountable, by virtue of $a_0$ being in $P$.
Thus every initial segment in $\powerset \N_{a_0}$ is countable, because it is not uncountable.
Then:
:$\Omega = \powerset \N_{a_0}$
{{qed}}
Theorem: Binomial Theorem/Hurwitz's Generalisation

:$\ds \paren {x + y}^n = \sum x \paren {x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n}^{\epsilon_1 + \cdots + \epsilon_n - 1} \paren {y - \epsilon_1 z_1 - \cdots - \epsilon_n z_n}^{n - \epsilon_1 - \cdots - \epsilon_n}$
where the summation ranges over all $2^n$ choices of $\epsilon_1, \ldots, \epsilon_n = 0$ or $1$ independently.

Proof:
Follows from this formula:
:$(1): \quad \ds \sum x \paren {x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n}^{\epsilon_1 + \cdots + \epsilon_n - 1} \paren {y + \paren {1 - \epsilon_1} z_1 - \cdots + \paren {1 - \epsilon_n} z_n}^{n - \epsilon_1 - \cdots - \epsilon_n} = \paren {x + y} \paren {x + y + z_1 + \cdots + z_n}^{n - 1}$
{{ProofWanted|... somehow}}
{{Namedfor|Adolf Hurwitz|cat = Hurwitz}}
Theorem: Set of Relations can be Ordered by Subset Relation

Let $S \times T$ be the product of two sets.
Let $\RR$ be a set of relations on $S \times T$.
Then $\RR$ can be ordered by the subset relation.

Proof:
Let $R$ be a relation on $S \times T$.
By the definition of relation, $R$ is associated with a subset $R \subseteq S \times T$.
Thus $\RR$ is a subset of the power set $\powerset {S \times T}$.
The result follows from Subset Relation is Ordering.
{{qed}}
Theorem: Set of Mappings can be Ordered by Subset Relation

Let $S \times T$ be the product of two sets.
Let $\FF$ be a set of mappings on $S \times T$.
Then $\FF$ can be ordered by the subset relation.

Proof:
By the definition of mapping, a mapping is a specific type of relation.
The result then follows from Set of Relations can be Ordered by Subset Relation.
{{qed}}
Category:Order Theory
Category:Subset Relation
j0y0swkto559bco1v1p98qjrul6xn9j
Theorem: Binomial Theorem/Abel's Generalisation/Negative n

Abel's Generalisation of Binomial Theorem:
{{:Abel's Generalisation of Binomial Theorem}}
does not hold for $n \in \Z_{< 0}$.

Proof:
Putting $n = x = -1$ and $y = z = 1$ into the {{LHS}}
:$\paren {-1 + 1}^{-1} = \dfrac 1 0$
which is undefined.
Putting the same values into the {{RHS}} gives:
{{begin-eqn}}
{{eqn | o = 
      | r = \sum_k \dbinom {-1} k \paren {-1} \paren {-1 - k}^{k - 1} \paren {1 + k}^{-1 - k}
      | c = 
}}
{{eqn | r = \sum_k \paren {-1}^k \dbinom {-1} k \paren {1 + k}^{-2}
      | c = 
}}
{{eqn | r = \sum_{k \mathop \ge 0} \paren {1 + k}^{-2}
      | c = 
}}
{{eqn | r = \sum_{k \mathop \ge 1} \dfrac 1 {k^2}
      | c = Translation of Index Variable of Summation: Corollary
}}
{{eqn | r = \dfrac {\pi^2} 6
      | c = Basel Problem
}}
{{end-eqn}}
{{qed}}
Theorem: Summation from k to m of r Choose k by s Choose n-k by nr-(r+s)k

:$\ds \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m + 1} \dbinom s {n - m}$

Proof:
The proof proceeds by induction over $m$.
For all $m \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
:$\ds \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m + 1} \dbinom s {n - m}$
Theorem: Factors of Binomial Coefficient/Corollary 1

For all $r \in \R, k \in \Z$:
:$\paren {r - k} \dbinom r k = r \dbinom {r - 1} k$
from which:
:$\dbinom r k = \dfrac r {r - k} \dbinom {r - 1} k$ (if $r \ne k$)

Proof:
{{begin-eqn}}
{{eqn | l = r \binom {r - 1} k
      | r = r \frac {\paren {r - 1} \paren {\paren {r - 1} - 1} \cdots \paren {\paren {r - 1} - k + 2} \paren {\paren {r - 1} - k + 1} } {k \paren {k - 1} \paren {k - 2} \cdots 1}
      | c = 
}}
{{eqn | r = \frac {r \paren {r - 1} \paren {r - 2} \cdots \paren {r - k + 1} \paren {r - k} } {k \paren {k - 1} \paren {k - 2} \cdots 1}
      | c = 
}}
{{eqn | r = \paren {r - k} \frac {r \paren {r - 1} \paren {r - 2} \cdots \paren {r - k + 1} } {k \paren {k - 1} \paren {k - 2} \cdots 1}
      | c = 
}}
{{eqn | r = \paren {r - k} \binom r k
      | c = 
}}
{{end-eqn}}
{{qed|lemma}}
Then:
:$\dbinom r k = \dfrac r {r - k} \dbinom {r - 1} k$
follows from the 
:$\paren {r - k} \dbinom r k = r \dbinom {r - 1} k$
by dividing both sides by $r - k$.
{{qed}}
Theorem: Factors of Binomial Coefficient/Corollary 2

For all $r \in \R, k \in \Z$:
:$\paren {r - k + 1} \dbinom r {k - 1} = k \dbinom r k$

Proof:
{{begin-eqn}}
{{eqn | l = \paren {r - k + 1} \dbinom r {k - 1}
      | r = \paren {r - k + 1} \frac {r \paren {r - 1} \paren {r - 2} \cdots \paren {r - \paren {k - 1} + 1} } {\paren {k - 1} \paren {k - 2} \cdots 1}
      | c = {{Defof|Binomial Coefficient|subdef = Real Numbers}}
}}
{{eqn | r = \frac {r \paren {r - 1} \paren {r - 2} \cdots \paren {r - k + 1} } {\paren {k - 1} \paren {k - 2} \cdots 1}
      | c = 
}}
{{eqn | r = k \times \frac {r \paren {r - 1} \paren {r - 2} \cdots \paren {r - k + 1} } {k \paren {k - 1} \paren {k - 2} \cdots 1}
}}
{{eqn | r = k \dbinom r k
      | c = {{Defof|Binomial Coefficient|subdef = Real Numbers}}
}}
{{end-eqn}}
{{qed}}
Category:Binomial Coefficients
k2gcvtasno1se9t2yhqnqazd1l9qjwy
Theorem: Inverse of Pascal's Triangle expressed as Matrix

Consider Pascal's triangle expressed as a (square) matrix $\mathbf M$, with the top left element holding $\dbinom 0 0$.
:$\begin{pmatrix}
1 &  0 &  0 &   0 &   0 &   0 &   0 &   0 &   0 & \cdots \\
1 &  1 &  0 &   0 &   0 &   0 &   0 &   0 &   0 & \cdots \\
1 &  2 &  1 &   0 &   0 &   0 &   0 &   0 &   0 & \cdots \\
1 &  3 &  3 &   1 &   0 &   0 &   0 &   0 &   0 & \cdots \\
1 &  4 &  6 &   4 &   1 &   0 &   0 &   0 &   0 & \cdots \\
1 &  5 & 10 &  10 &   5 &   1 &   0 &   0 &   0 & \cdots \\
1 &  6 & 15 &  20 &  15 &   6 &   1 &   0 &   0 & \cdots \\
1 &  7 & 21 &  35 &  35 &  21 &   7 &   1 &   0 & \cdots \\
1 &  8 & 28 &  56 &  70 &  56 &  28 &   8 &   1 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\
\end{pmatrix}$
The inverse $\mathbf M^{-1}$ of $\mathbf M$ is the same as $\mathbf M$ but with alternate elements negated, starting with the elements below the main diagonal:
:$\begin{pmatrix}
 1 &  0 &   0 &   0 &   0 &   0 &  0 &  0 & 0 & \cdots \\
-1 &  1 &   0 &   0 &   0 &   0 &  0 &  0 & 0 & \cdots \\
 1 & -2 &   1 &   0 &   0 &   0 &  0 &  0 & 0 & \cdots \\
-1 &  3 &  -3 &   1 &   0 &   0 &  0 &  0 & 0 & \cdots \\
 1 & -4 &   6 &  -4 &   1 &   0 &  0 &  0 & 0 & \cdots \\
-1 &  5 & -10 &  10 &  -5 &   1 &  0 &  0 & 0 & \cdots \\
 1 & -6 &  15 & -20 &  15 &  -6 &  1 &  0 & 0 & \cdots \\
-1 &  7 & -21 &  35 & -35 &  21 & -7 &  1 & 0 & \cdots \\
 1 & -8 &  28 & -56 &  70 & -56 & 28 & -8 & 1 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\
\end{pmatrix}$

Proof:
We have from Sum over $k$ of $\dbinom r k$ by $\dbinom {s + k} n$ by $\left({-1}\right)^{r - k}$: Corollary:
{{:Sum over k of r Choose k by s+k Choose n by -1^r-k/Corollary}}
By definition of matrix multiplication, this is the element $a_{r n}$ of the matrix formed by multiplying the two matrices above.
As can be seen, this results in the identity matrix .
{{qed}}
Theorem: Binomial Coefficient of Half

Let $k \in \Z$.
:$\dbinom {\frac 1 2} k = \dfrac {\left({-1}\right)^{k - 1} } {4^k \left({2 k - 1}\right)} \dbinom {2 k} k$
where $\dbinom {\frac 1 2} k$ denotes a binomial coefficient.

Proof:
{{begin-eqn}}
{{eqn | l = \dbinom {\frac 1 2} k
      | r = \dfrac {1/2} {1/2 - k} \dbinom {1/2 - 1} k
      | c = Factors of Binomial Coefficient: Corollary 1
}}
{{eqn | r = \dfrac 1 {1 - 2 k} \dbinom {-1/2} k
      | c = 
}}
{{eqn | r = \dfrac 1 {1 - 2 k} \dfrac {\left({-1}\right)^k} {4^k} \dbinom {2 k} k
      | c = Binomial Coefficient of Minus Half
}}
{{eqn | r = \dfrac {\left({-1}\right)^{k - 1} } {4^k \left({2 k - 1}\right)} \dbinom {2 k} k
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Binomial Coefficient of Half/Corollary

Let $k \in \Z_{\ge 0}$.
:$\dbinom {\frac 1 2} k = \dfrac {\paren {-1}^{k - 1} } {2^{2 k - 1} \paren {2 k - 1} } \dbinom {2 k - 1} k - \delta_{k 0}$
where:
:$\dbinom {\frac 1 2} k$ denotes a binomial coefficient
:$\delta_{k 0}$ denotes the Kronecker delta.

Proof:
When $k > 0$ we have:
{{begin-eqn}}
{{eqn | l = \dfrac {\paren {-1}^{k - 1} } {4^k \paren {2 k - 1} } \dbinom {2 k} k
      | r = \dfrac {\paren {-1}^{k - 1} } {4^k \paren {2 k - 1} } \dfrac {2 k} {2 k - k} \dbinom {2 k - 1} k
      | c = {{Corollary|Factors of Binomial Coefficient|1}}
}}
{{eqn | r = \dfrac {\paren {-1}^{k - 1} } {4^k \paren {2 k - 1} } \dfrac {2 k} k \dbinom {2 k - 1} k
      | c = 
}}
{{eqn | r = \dfrac {2 \paren {-1}^{k - 1} } {4^k \paren {2 k - 1} } \dbinom {2 k - 1} k
      | c = 
}}
{{end-eqn}}
When $k = 0$ we have:
{{begin-eqn}}
{{eqn | l = \dfrac {\paren {-1}^{k - 1} } {4^k \paren {2 k - 1} } \dbinom {2 k} k
      | r = \dfrac {\paren {-1}^{-1} } {-1} \dbinom 0 0
      | c = 
}}
{{eqn | r = 1
      | c = 
}}
{{end-eqn}}
while:
{{begin-eqn}}
{{eqn | l = \dfrac {\paren {-1}^{k - 1} } {2^{2 k - 1} \paren {2 k - 1} } \dbinom {2 k - 1} k
      | r = \dfrac {\paren {-1}^{-1} } {2^{-1} \paren {-1} } \dbinom {-1} 0
      | c = 
}}
{{eqn | r = \dfrac 1 {2^{-1} } \dbinom {-1} 0
      | c = 
}}
{{eqn | r = 2 \dbinom {-1} 0
      | c = 
}}
{{eqn | r = 2 \paren {-1}^0 \dbinom {0 - \paren {-1} - 1} 0
      | c = Negated Upper Index of Binomial Coefficient
}}
{{eqn | r = 2 \dbinom 0 0
      | c = 
}}
{{eqn | r = 2
      | c = 
}}
{{end-eqn}}
Hence:
:$\dbinom {\frac 1 2} k = \dfrac {2 \paren {-1}^{k - 1} } {4^k \paren {2 k - 1} } \dbinom {2 k - 1} k - \delta_{k 0}$
{{qed}}
Theorem: Inverse of Stirling's Triangle expressed as Matrix

Consider Stirling's triangle of the first kind (signed) expressed as a (square) matrix $\mathbf A$, with the top left element holding $\map s {0, 0}$.
:$\begin{pmatrix}
1 &     0 &       0 &      0 &      0 &     0 & \cdots \\
0 &     1 &       0 &      0 &      0 &     0 & \cdots \\
0 &    -1 &       1 &      0 &      0 &     0 & \cdots \\
0 &     2 &      -3 &      1 &      0 &     0 & \cdots \\
0 &    -6 &      11 &     -6 &      1 &     0 & \cdots \\
0 &    24 &     -50 &     35 &    -10 &     1 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots &  \ddots \\
\end{pmatrix}$
Then consider Stirling's triangle of the second kind expressed as a (square) matrix $\mathbf B$, with the top left element holding $\ds {0 \brace 0}$.
:$\begin{pmatrix}
1 & 0 &   0 &    0 &    0 &    0 & \cdots \\
0 & 1 &   0 &    0 &    0 &    0 & \cdots \\
0 & 1 &   1 &    0 &    0 &    0 & \cdots \\
0 & 1 &   3 &    1 &    0 &    0 & \cdots \\
0 & 1 &   7 &    6 &    1 &    0 & \cdots \\
0 & 1 &  15 &   25 &   10 &    1 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\
\end{pmatrix}$
Then:
:$\mathbf A = \mathbf B^{-1}$
that is:
:$\mathbf B = \mathbf A^{-1}$

Proof:
First note that from Relation between Signed and Unsigned Stirling Numbers of the First Kind:
:$\ds {n \brack m} = \paren {-1}^{n + m} \map s {n, m}$
From First Inversion Formula for Stirling Numbers:
:$\ds \sum_k {n \brack k} {k \brace m} \paren {-1}^{n - k} = \delta_{m n}$
From Second Inversion Formula for Stirling Numbers:
:$\ds \sum_k {n \brace k} {k \brack m} \paren {-1}^{n - k} = \delta_{m n}$
By definition of matrix multiplication, the element $a_{r n}$ of the matrix formed by multiplying the two matrices above.
As can be seen, this results in the identity matrix .
{{qed}}
Theorem: Wosets are Isomorphic to Each Other or Initial Segments

Let $\struct {S, \preceq_S}$ and $\struct {T, \preceq_T}$ be well-ordered sets.
Then precisely one of the following hold:
:$\struct {S, \preceq_S}$ is order isomorphic to $\struct {T, \preceq_T}$
or:
:$\struct {S, \preceq_S}$ is order isomorphic to an initial segment in $\struct {T, \preceq_T}$
or:
:$\struct {T, \preceq_T}$ is order isomorphic to an initial segment in $\struct {S, \preceq_S}$

Proof:
We assume $S \ne \O \ne T$; otherwise the theorem holds vacuously. 
Define:
:$S' = S \cup \text{ initial segments in } S$
:$T' = T \cup \text{ initial segments in } T$
:$\FF = \set {f: S' \to T' \mid f \text{ is an order isomorphism} }$
We note that $\FF$ is non-empty, because it at least contains a trivial order isomorphism between singletons:
:$f_0: \set {\text{smallest element in } S} \to \set {\text{smallest element in } T}$
Such smallest elements are guaranteed to exist by virtue of $S$ and $T$ being well-ordered.
By the definition of initial segment, the initial segments of $S$ are subsets of $S$.
For any initial segment $I_\alpha$ of $S$, such a segment has an upper bound by definition, namely, $\alpha$.
Also, no initial segment of $S'$ is the entirety of $S$.
Thus every initial segment of $S'$ has an upper bound, $S$ itself.
Every chain in $S'$ has an upper bound, because it defines an initial segment.  
The previous reasoning also applies to $T'$.
By Antilexicographic Product of Totally Ordered Sets is Totally Ordered, $S' \times T'$ is itself a totally ordered set, with upper bound $S \times T$.
Thus the hypotheses of Zorn's Lemma are satisfied for $\FF$.
Let $f_1$ be a maximal element of $\FF$.
Call its domain $A$ and its codomain $B$.
Suppose $A$ is an initial segment $I_a$ in $S$ and $B$ is an initial segment $I_b$ in $T$.
Then $f_1$ can be extended by defining $\map {f_1} a = b$.
This would contradict $f_1$ being maximal, so it cannot be the case that both $A$ and $B$ are initial segments.
Then precisely one of the following hold:
:$A = S$, with $S$ order isomorphic to an initial segment in $T$
or:
:$B = T$, with $T$ order isomorphic to an initial segment in $S$
or:
:both $A = S$ and $B = T$, with $S$ order isomorphic to $T$.
The cases are distinct by Well-Ordered Class is not Isomorphic to Initial Segment.
{{qed}}
Theorem: Minimal Uncountable Well-Ordered Set Unique up to Isomorphism

Let $\Omega, \Omega'$ be minimal uncountable well-ordered sets.
Then $\Omega$ is order isomorphic to $\Omega'$.
That is, the minimal uncountable well-ordered set is unique up to order isomorphism.

Proof:
From Wosets are Isomorphic to Each Other or Initial Segments, precisely one of the following holds:
:$\Omega$ is order isomorphic to $\Omega'$
or:
:$\Omega$ is order isomorphic to an initial segment in $\Omega'$
or:
:$\Omega'$ is order isomorphic to an initial segment in $\Omega$.
By the definition of minimal uncountable well-ordered set, $\Omega$ and $\Omega'$ are uncountable.
The initial segments of $\Omega$ and $\Omega'$ are countable.
An uncountable set cannot be order isomorphic to a countable set.
{{MissingLinks|the above claim needs proof. The theorem might or might not exist already on the wiki}}
Thus $\Omega$ must be order isomorphic to $\Omega'$.
The result follows from the definition of uniqueness.
{{qed}}
Theorem: Addition Rule for Gaussian Binomial Coefficients/Formulation 1

:$\dbinom n m_q = \dbinom {n - 1} m_q + \dbinom {n - 1} {m - 1}_q q^{n - m}$

Proof:
By definition of Gaussian binomial coefficient:
{{begin-eqn}}
{{eqn | l = \binom n m_q
      | o = :=
      | r = \prod_{j \mathop = 0}^{m - 1} \dfrac {1 - q^{n - j} } {1 - q^{j + 1} }
      | c = 
}}
{{eqn | r = \dfrac {\paren {1 - q^n} \paren {1 - q^{n - 1} } \dotsm \paren {1 - q^{n - m + 1} } } {\paren {1 - q^m} \paren {1 - q^{m - 1} } \dotsm \paren {1 - q^1} }
      | c = 
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \dbinom {n - 1} m_q + \dbinom {n - 1} {m - 1}_q q^{n - m}
      | r = \prod_{j \mathop = 0}^{m - 1} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } + q^{n - m} \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} }
      | c = {{Defof|Gaussian Binomial Coefficient}}
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } \paren {\dfrac {1 - q^{n - 1 - \paren {m - 1} } } {1 - q^{\paren {m - 1} + 1} } + q^{n - m} }
      | c = extracting $\ds \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} }$ as a factor
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } \paren {\dfrac {1 - q^{n - m} } {1 - q^m} + q^{n - m} }
      | c = simplifying
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } \paren {\dfrac {1 - q^{n - m} + q^{n - m} \paren {1 - q^m} } {1 - q^m} }
      | c = placing over a common denominator
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } \paren {\dfrac {1 - q^{n - m} + q^{n - m} - q^{n - m} q^m} {1 - q^m} }
      | c = 
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } \paren {\dfrac {1 - q^n} {1 - q^m} }
      | c = 
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 1} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} }
      | c = 
}}
{{eqn | r = \binom n m_q
      | c = {{Defof|Gaussian Binomial Coefficient}}
}}
{{end-eqn}}
{{qed}}
Theorem: Addition Rule for Gaussian Binomial Coefficients/Formulation 2

:$\dbinom n m_q = \dbinom {n - 1} m_q q^m + \dbinom {n - 1} {m - 1}_q$

Proof:
By definition of Gaussian binomial coefficient:
{{begin-eqn}}
{{eqn | l = \binom n m_q
      | o = :=
      | r = \prod_{j \mathop = 0}^{m - 1} \dfrac {1 - q^{n - j} } {1 - q^{j + 1} }
      | c = 
}}
{{eqn | r = \dfrac {\paren {1 - q^n} \paren {1 - q^{n - 1} } \dotsm \paren {1 - q^{n - m + 1} } } {\paren {1 - q^m} \paren {1 - q^{m - 1} } \dotsm \paren {1 - q^1} }
      | c = 
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \dbinom {n - 1} m_q q^m + \dbinom {n - 1} {m - 1}_q
      | r = q^m \prod_{j \mathop = 0}^{m - 1} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } + \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} }
      | c = {{Defof|Gaussian Binomial Coefficient}}
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } \paren {q^m \dfrac {1 - q^{n - 1 - \paren {m - 1} } } {1 - q^{\paren {m - 1} + 1} } + 1}
      | c = extracting $\ds \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} }$ as a factor
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } \paren {q^m \dfrac {1 - q^{n - m} } {1 - q^m} + 1}
      | c = simplifying
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } \paren {\dfrac {q^m \paren {1 - q^{n - m} } + \paren {1 - q^m} } {1 - q^m} }
      | c = placing over a common denominator
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } \paren {\dfrac {q^m - q^n + 1 - q^m} {1 - q^m} }
      | c = 
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } \paren {\dfrac {1 - q^n} {1 - q^m} }
      | c = 
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 1} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} }
      | c = 
}}
{{eqn | r = \binom n m_q
      | c = {{Defof|Gaussian Binomial Coefficient}}
}}
{{end-eqn}}
{{qed}}
Theorem: Union of Initial Segments is Initial Segment or All of Woset

Let $\struct {X, \preccurlyeq}$ be a well-ordered non-empty set.
Let $A \subseteq X$.
Let:
:$\ds J = \bigcup_{x \mathop \in A} S_x$
be a union of initial segments defined by the elements of $A$.
Then either:
:$J = X$
or:
:$J$ is an initial segment of $X$.

Proof:
Suppose the hypotheses of the theorem hold.
If $J = X$ then the theorem is satisfied.
Assume $J \ne X$.
Then $X \setminus J$ is non-empty.
By Subset of Well-Ordered Set is Well-Ordered, $X \setminus J$ is itself well-ordered.
Thus $X \setminus J$ has a smallest element; call it $b$.
We claim that $b \preccurlyeq y$ for every $y \in J$.
{{AimForCont}} there exists a $y \in J$ with $b \preccurlyeq y$.
Then there exists some $x_0 \in A$ with $y$ in the initial segment $S_{x_0}$.
Thus $b \in S_{x_0}$ and so $b \in J$.
This contradicts the fact that $b \in X \setminus J$.
Thus there cannot exist a $y \in J$ with $b \preccurlyeq y$.
Any element strictly preceding $b$ is in $J$, by choice of $b$.
Also, $b \notin J$, because $b \in X \setminus J$.
From above, any element strictly succeeding $b$ is not in $J$.
Thus $J = S_b$ by definition of initial segment.
{{qed}}
Theorem: Union of Initial Segments is Initial Segment or All of Woset

Let $\struct {X, \preccurlyeq}$ be a well-ordered non-empty set.
Let $A \subseteq X$.
Let:
:$\ds J = \bigcup_{x \mathop \in A} S_x$
be a union of initial segments defined by the elements of $A$.
Then either:
:$J = X$
or:
:$J$ is an initial segment of $X$.

Proof:
Suppose the hypotheses of the theorem hold.
For every $x \in A$, we have $S_x \subseteq X$ by the definition of initial segment.
By Union of Family of Subsets is Subset, we have $J \subseteq X$.
Consider $X \setminus J$.
If $X \setminus J$ is empty, then $X \subseteq J$ by Set Difference with Superset is Empty Set, and hence $J = X$ by Subset Relation is Antisymmetric, so the theorem is satisfied.
So assume $X \setminus J$ is non-empty.
By Subset of Well-Ordered Set is Well-Ordered, $X \setminus J$ is itself well-ordered.
Thus $X \setminus J$ has a smallest element; call it $a$.
We will show that $J = S_a$, thus showing the veracity of the theorem.
We need to show that:
:$J = \set {b \in S: b \prec a}$
which by definition of set equality is equivalent to the claim that:
:$\forall z \in X:  \in J \iff z \prec a$
For the forward direction, suppose $z \in J$.
We need to show that $z \prec a$.
By definition of union, there is $x \in A$ such that $z \in S_x$, that is:
:$z \prec x$
{{AimForCont}} $a \preceq z$.
Then $a \prec x$, so $a \in S_x$ by the definition of initial segment
Hence $a \in J$ by definition of union.
This contradicts the fact that $a \in X \setminus J$.
Thus $z \prec a$.
For the backward direction, suppose $z \prec a$ where $z \in X$. We need to show that $z \in J$
{{AimForCont}} $z \notin J$.
Then $z \in X \setminus J$.
Hence $a \preceq z$ by definition of smallest element.
This contradicts the fact that $z \prec a$.
Thus $z \in J$.
Since $z$ was arbitrary, we conclude that $J = S_a$ as required.
{{qed}}
Theorem: Countable Subset of Minimal Uncountable Well-Ordered Set Has Upper Bound

Let $\Omega$ denote the minimal uncountable well-ordered set.
Let $\omega$ be a countable subset of $\Omega$.
Then $\omega$ has an upper bound in $\Omega$.

Proof:
By the minimal uncountable well-ordered set, $\Omega$ is uncountable
Then $\omega \ne \Omega$, for the former is countable and the latter is not.
Consider the union:
:$\ds \bigcup_{x \mathop \in \omega} S_x$
of initial segments $S_x$ in $\omega$.
By the definition of $\Omega$, any $S_x$ is countable.
Thus $\ds \bigcup_{x \mathop \in \omega} S_x$ is a countable union of countable sets.
This union is also countable, and so also cannot be all of $\Omega$, as $\Omega$ is uncountable.
From Union of Initial Segments is Initial Segment or All of Woset, there exists some $y \in \Omega$ such that:
:$\ds \bigcup_{x \mathop \in \omega} S_x = S_y$
Such a $y$ is an upper bound for $\omega$ by the definition of an initial segment.
{{qed}}
{{AoC|Countable Union of Countable Sets is Countable}}