nyuuzyou/wikis huggingface dataset:
Second Order ODE/y y'' = y^2 y' + (y')^2

== Theorem ==
The second order ODE:
:$(1): \quad y y = y^2 y' + \paren {y'}^2$
subject to the initial conditions:
:$y = -\dfrac 1 2$ and $y' = 1$ when $x = 0$
has the particular solution:
:$2 y - 3 = 8 y \, \map \exp {\dfrac {3 x} 2}$
== Proof ==
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as:
Now to consider the initial conditions:
:$y = -\dfrac 1 2$ and $y' = 1$ when $x = 0$
After algebra:
When $x = 0$ we have $y = -1/2$:
Differentiating to get $y'$:
:$y' \paren {1 - 2 C_1} = \paren {y + C_1} C_1 e^{C_1 x} + e^{C_1 x} y'$
Putting $y' = 1$ when $x = 0$ we get:
So:
== Sources ==
* : $\S 2.11$: Problem $2 \ \text{(b)}$
Category:Examples of Second Order ODEs
Second Order Weakly Stationary Gaussian Stochastic Process is Strictly Stationary

== Theorem ==
Let $S$ be a Gaussian stochastic process giving rise to a time series $T$.
Let $S$ be weakly stationary of order $2$.
Then $S$ is strictly stationary.
== Proof ==
By definition of a Gaussian process, the probability distribution of $T$ be a multivariate normal distribution.
By definition, a normal distribution is characterized completely by its expectation and its variance.
That is, its $1$st and $2$nd moments.
The result follows.
== Sources ==
* :
::Part $\text {I}$: Stochastic Models and their Forecasting:
:::$2$: Autocorrelation Function and Spectrum of Stationary Processes:
::::$2.1$ Autocorrelation Properties of Stationary Models:
:::::$2.1.3$ Positive Definiteness and the Autocovariance Matrix: Weak stationarity
Category:Stationary Stochastic Processes
Second Pappus-Guldinus Theorem

== Theorem ==
Let $C$ be a plane figure that lies entirely on one side of a straight line $L$.
Let $S$ be the solid of revolution generated by $C$ around $L$.
Then the surface area of $S$ is equal to the perimeter length of $C$ multiplied by the distance travelled by the centroid of $C$ around $L$ when generating $S$.
== Proof ==
== Also known as ==
== Also see ==
* First Pappus-Guldinus Theorem
== Sources ==
* 
* 
* 
Category:Pappus's Theorems
Category:Surfaces of Revolution
Second Partial Derivative/Examples/u + ln u = x y

== Examples of Second Partial Derivative ==
Let $u + \ln u = x y$ be an implicit function.
Then:
:$\dfrac {\partial^2 u} {\partial y \partial x} = \dfrac {\partial^2 u} {\partial x \partial y} = \dfrac u {u + 1} + \dfrac {x y u} {\paren {u + 1}^2}$
== Proof ==
From Partial Derivative Examples: u + ln u = x y:
Then:
Then we have that:
:$\dfrac {\partial^2 u} {\partial x \partial y} = \map {\dfrac \partial {\partial x} } {\dfrac {u x} {u + 1} }$
which is exactly the same as:
:$\dfrac {\partial^2 u} {\partial y \partial x} = \map {\dfrac \partial {\partial y} } {\dfrac {u y} {u + 1} }$
but with $y$ and $x$ exchanged.
Hence:
and it is seen that the second partial derivatives are indeed the same.
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: $1.3$ Higher Order Derivatives: Example $\text E$
Category:Examples of Partial Derivatives
Second Partial Derivative wrt r of ln (r^2 + s)

== Theorem ==
:$\dfrac {\partial^2} {\partial r^2} \map \ln {r^2 + s} = \dfrac {2 \paren {s - r^2} } {\paren {r^2 + s}^2}$
== Proof ==
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $2$
Category:Examples of Partial Derivatives
Second Pentagonal Number/Sequence

== Sequence of Second Pentagonal Numbers ==
The sequence of second pentagonal numbers, for $n \in \Z_{\ge 0}$, begins:
:$0, 2, 7, 15, 26, 40, 57, 77, 100, 126, 155, 187, \ldots$
== Sources ==
* : $22$
* : $22$
Category:Second Pentagonal Numbers
Second Price Auction/Analysis

== Analysis of Second Price Auction ==
Let $G$ denoted the game under analysis, that is, the second price auction under discussion.
Let $F$ be the object being bid for.
As in the formal definition of the sealed-bid auction, let the players be labelled in order of their valuations of $F$:
:$v_1 > v_2 > \cdots > v_n > 0$
In this context, a move by player $i$ is the bid that $i$ places on $F$.
Let $b$ be the profile of moves made by all players.
Let $\map m b$ be defined as the lowest $j$ such that $\ds b_j = \max_{k \mathop \in \set {1, 2, \ldots, n} } b_k$.
That is, $\map m b$ is the index of the player who has the lowest index of all players who place the highest bid on $F$.
Let player $i$ win the auction by move $b_i$.
By hypothesis, player $i$'s valuation for $F$ is $v_i$.
Hence $i$'s payoff is $v_i - \max_{j \mathop \ne 1} b_j$.
For the other players the payoff is $0$.
Thus the payoff function of player $i$ is defined as follows:
:$\map {p_i} b = \begin{cases} v_i − \max_{j \mathop \ne i} b_j & : i = \map m b \\ 0 & : \text{otherwise} \end{cases}$
$b$ is a Nash equilibrium :
:$\ds \max_{j \mathop \ne i} v_j \le \max_{j \mathop \ne i} b_j = b_i \le v_1$
That is:
:$(1): \quad \ds \max_{j \mathop \ne i} v_j \le b_i$ (that is, player $i$'s bid was high enough to win)
:$(2): \quad \ds \max_{j \mathop \ne i} b_j \le v_i$ (that is, player $i$'s valuation was high enough to avoid suffering the winner's curse).
== Proof ==
=== Sufficient Condition ===
Suppose that $\ds \max_{j \mathop \ne i} v_j > b_i$.
Then player $j$ such that $v_j > b_i$ can win $F$ by submitting a bid in the open interval $\openint {b_i} {v_j}$, say $v_j − \epsilon$.
Then his payoff increases from $0$ to $v_j - b_i$.
Thus if $\ds \max_{j \mathop \ne i} v_j > b_i$, $b$ is not a Nash equilibrium.
Suppose $\ds \max_{j \mathop \ne i} b_j > v_i$.
Then player $i$ has a negative payoff.
$i$ can then increase that payoff to $0$ by submitting a losing bid.
Thus if $\ds \max_{j \mathop \ne i} b_j > v_i$, $b$ is not a Nash equilibrium.
Thus it can be seen that if any of the conditions $(1)$ to $(3)$ do not hold, then $b$ is not a Nash equilibrium.
=== Necessary Condition ===
Let $b$ be a profile of moves that satisfies $(1)$ and $(2)$.
Thus player $i$ is the winner.
By $(2)$, his payoff is non-negative.
By submitting another bid either:
:$i$ remains the winner with the same payoff
or:
:$i$ becomes a loser, with a payoff of $0$.
The payoff of any other player $j$ is $0$.
This can increase only if $j$ bids more and becomes the winner.
Then his payoff is $v_j - b_i$.
But by $(1)$, this payoff is negative.
So $b$ is a Nash equilibrium.
== Also see ==
* Definition:Winner's Curse: when the valuation is less than the bid, and the payoff is thereby negative.
== Sources ==
* : $\text I$ Strategic Games: Chapter $2$ Nash Equilibrium: $2.3$: Examples: Exercise $18.3$
Category:Auctions
Second Principle of Finite Induction

== Theorem ==
Let $S \subseteq \Z$ be a subset of the integers.
Let $n_0 \in \Z$ be given.
Suppose that:
:$(1): \quad n_0 \in S$
:$(2): \quad \forall n \ge n_0: \paren {\forall k: n_0 \le k \le n \implies k \in S} \implies n + 1 \in S$
Then:
:$\forall n \ge n_0: n \in S$
The second principle of finite induction is usually stated and demonstrated for $n_0$ being either $0$ or $1$.
This is often dependent upon whether the analysis of the fundamentals of mathematical logic are zero-based or one-based.
=== Zero-Based ===
=== One-Based ===
== Proof ==
Define $T$ as:
:$T = \set {n \in \Z : \forall k: n_0 \le k \le n: k \in S}$
Since $n \le n$, it follows that $T \subseteq S$.
Therefore, it will suffice to show that:
:$\forall n \ge n_0: n \in T$
Firstly, we have that $n_0 \in T$ the following condition holds:
:$\forall k: n_0 \le k \le n_0 \implies k \in S$
Since $n_0 \in S$, it thus follows that $n_0 \in T$.
Now suppose that $n \in T$; that is:
:$\forall k: n_0 \le k \le n \implies k \in S$
By $(2)$, this implies:
:$n + 1 \in S$
Thus, we have:
:$\forall k: n_0 \le k \le n + 1 \implies k \in S$
Therefore, $n + 1 \in T$.
Hence, by the Principle of Finite Induction:
:$\forall n \ge n_0: n \in T$
as desired.
== Terminology ==
== Also known as ==
== Also see ==
* Principle of Mathematical Induction
* Principle of Finite Induction
* Second Principle of Mathematical Induction
* Equivalence of Well-Ordering Principle and Induction
== Algorithmic Proof Procedure ==
Category:Named Theorems
Category:Principle of Finite Induction
Category:Mathematical Induction
Category:Proof Techniques
Category:Second Principle of Finite Induction
Second Principle of Finite Induction/One-Based

== Theorem ==
Let $S \subseteq \N_{>0}$ be a subset of the $1$-based natural numbers.
Suppose that:
:$(1): \quad 1 \in S$
:$(2): \quad \forall n \in \N_{>0}: \paren {\forall k: 1 \le k \le n \implies k \in S} \implies n + 1 \in S$
Then:
:$S = \N_{>0}$
== Proof ==
Define $T$ as:
:$T = \set {n \in \N_{>0}: \forall k: 1 \le k \le n: k \in S}$
Since $n \le n$, it follows that $T \subseteq S$.
Therefore, it will suffice to show that:
:$\forall n \ge 1: n \in T$
Firstly, we have that $1 \in T$ the following condition holds:
:$\forall k: 1 \le k \le 1 \implies k \in S$
Since $1 \in S$, it thus follows that $1 \in T$.
Now suppose that $n \in T$; that is:
:$\forall k: 1 \le k \le n \implies k \in S$
By $(2)$, this implies:
:$n + 1 \in S$
Thus, we have:
:$\forall k: 1 \le k \le n + 1 \implies k \in S$
Therefore, $n + 1 \in T$.
Hence, by the Principle of Finite Induction:
:$\forall n \ge 1: n \in T$
That is:
:$T = \N_{>0}$
and as $S \subseteq \N_{>0}$ it follows that:
:$S = N_{>0}$
== Sources ==
* : Chapter $1$: Properties of the Natural Numbers: $\S 20 \alpha$
* : Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: $\mathbf{I}'$
Category:Second Principle of Finite Induction
Second Principle of Finite Induction/Zero-Based

== Theorem ==
Let $S \subseteq \N$ be a subset of the natural numbers.
Suppose that:
:$(1): \quad 0 \in S$
:$(2): \quad \forall n \in \N: \paren {\forall k: 0 \le k \le n \implies k \in S} \implies n + 1 \in S$
Then:
:$S = \N$
== Proof ==
Define $T$ as:
:$T = \set {n \in \N : \forall k: 0 \le k \le n: k \in S}$
Since $n \le n$, it follows that $T \subseteq S$.
Therefore, it will suffice to show that:
:$\forall n \ge 0: n \in T$
Firstly, we have that $0 \in T$ the following condition holds:
:$\forall k: 0 \le k \le 0 \implies k \in S$
Since $0 \in S$, it thus follows that $0 \in T$.
Now suppose that $n \in T$; that is:
:$\forall k: 0 \le k \le n \implies k \in S$
By $(2)$, this implies:
:$n + 1 \in S$
Thus, we have:
:$\forall k: 0 \le k \le n + 1 \implies k \in S$
Therefore, $n + 1 \in T$.
Hence, by the Principle of Finite Induction:
:$\forall n \ge 0: n \in T$
That is:
:$T = \N$
and as $S \subseteq \N$ it follows that:
:$S = N$
Category:Second Principle of Finite Induction
Second Principle of Mathematical Induction

== Theorem ==
Let $\map P n$ be a propositional function depending on $n \in \Z$.
Let $n_0 \in \Z$ be given.
Suppose that:
:$(1): \quad \map P {n_0}$ is true
:$(2): \quad \forall k \in \Z: k \ge n_0: \map P {n_0} \land \map P {n_0 + 1} \land \ldots \land \map P {k - 1} \land \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is true for all $n \ge n_0$.
This process is called proof by (mathematical) induction.
The Second Principle of Mathematical Induction is usually stated and demonstrated for $n_0$ being either $0$ or $1$.
This is often dependent upon whether the analysis of the fundamentals of mathematical logic are zero-based or one-based.
=== Zero-Based ===
=== One-Based ===
== Proof ==
For each $n \ge n_0$, let $\map {P'} n$ be defined as:
:$\map {P'} n := \map P {n_0} \land \dots \land \map P n$
It suffices to show that $\map {P'} n$ is true for all $n \ge n_0$.
It is immediate from the assumption $\map P {n_0}$ that $\map {P'} {n_0}$ is true.
Now suppose that $\map {P'} n$ holds.
By $(2)$, this implies that $\map P {n + 1}$ holds as well.
Consequently, $\map {P'} n \land \map P {n + 1} = \map {P'} {n + 1}$ holds.
Thus by the Principle of Mathematical Induction:
:$\map {P'} n$ holds for all $n \ge n_0$
as desired.
== Also defined as ==
== Terminology ==
== Also known as ==
== Also see ==
* Principle of Mathematical Induction
* Principle of Finite Induction
* Second Principle of Finite Induction
== Sources ==
* : $\S 0.1$. Arithmetic
* : Appendix $\text{A}.6$: Mathematical Induction: Theorem $\text{A}.8$
* 
* 
* : Chapter $3$: The Natural Numbers: $\S 3$ Derivation of the Peano postulates and other results: Exercise $3.1$
Category:Second Principle of Mathematical Induction
Category:Principle of Mathematical Induction
Category:Mathematical Induction
Category:Proof Techniques
Second Principle of Mathematical Induction/Algorithm

== Algorithm ==
The Second Principle of Mathematical Induction can be implemented as an algorithm as follows.
Let $\map P n$ be a propositional function depending on $n \in \N$.
Let it be established that:
:$\text{(a)}: \quad \map P 1$ is true
:$\text{(b)}: \quad$ If all of $\map P 1, \map P 2, \ldots, \map P k$ are true, then $\map P {k + 1}$ is true.
Then the following algorithm provides a proof of $\map P k$ for all $n \in \N$:
:$\mathbf 1:$ Prove $\map P 1$.
::Set $k \gets 1$ and prove $\map P 1$ according to $\text{(a)}$.
:$\mathbf 2:$ Is $k = n$?.
::If $k = n$ then terminate. The required proof has been output.
:$\mathbf 3:$ Prove $\map P {k + 1}$.
::According to $\text{(b)}$, output a proof that if all of $\map P 1, \map P 2, \ldots, \map P k$ are true, then $\map P {k + 1}$ is true.
:$\mathbf 4:$ Increase $k$.
::$k \gets k + 1$ and to to step $\mathbf 2$.
== Sources ==
* : $\S 1.2.1$: Mathematical Induction
Category:Second Principle of Mathematical Induction
Second Principle of Mathematical Induction/Also defined as

== Second Principle of Mathematical Induction: Also known as == 
The Second Principle of Mathematical Induction can often be found stated more informally, inasmuch as the propositional function $P$ is referred to as "a statement about integers".
== Sources ==
* : $\S 0.1$. Arithmetic
* 
* 
Category:Principle of Mathematical Induction
Second Principle of Mathematical Induction/One-Based

== Theorem ==
Let $\map P n$ be a propositional function depending on $n \in \N_{>0}$.
Suppose that:
:$(1): \quad \map P 1$ is true
:$(2): \quad \forall k \in \N_{>0}: \map P 1 \land \map P 2 \land \ldots \land \map P {k - 1} \land \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is true for all $n \in \N_{>0}$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
== Sources ==
* : $\text {1-1}$ Principle of Mathematical Induction
* : $\S 1.2.1$: Mathematical Induction
* : $\S 1.4.2$: Mathematical induction
Category:Second Principle of Mathematical Induction
Second Principle of Mathematical Induction/One-Based/Proof 1

== Theorem ==
== Proof ==
For each $n \in \N_{>0}$, let $\map {P'} n$ be defined as:
:$\map {P'} n := \map P 1 \land \dots \land \map P n$
It suffices to show that $\map {P'} n$ is true for all $n \in \N_{>0}$.
It is immediate from the assumption $\map P 1$ that $\map {P'} 1$ is true.
Now suppose that $\map {P'} n$ holds.
By $(2)$, this implies that $\map P {n + 1}$ holds as well.
Consequently, $\map {P'} n \land \map P {n + 1} = \map {P'} {n + 1}$ holds.
Thus by the Principle of Mathematical Induction:
:$\map {P'} n$ holds for all $n \in \N_{>0}$
as desired.
Category:Second Principle of Mathematical Induction
Second Principle of Mathematical Induction/One-Based/Proof 2

== Theorem ==
== Proof ==
Let $S \subseteq \N_{>0}$ containing those $n \in \N_{>0}$ for which $\map P n$ does not hold.
 $S \ne \O$.
Then by the Well-Ordering Principle $S$ contains a minimal element $s$.
We have that $s \ne 1$ because $\map P 1$ is true from $(1)$.
Thus there must exist some $k \in \N_{>0}$ such that $s = k + 1$.
As $k + 1$ is the minimal element of $S$ it follows that all $n$ such that $n < k + 1 = s$ are not in $S$
But this means that $\map P n$ is true for all $n < s$
But by $(2)$ it follows that $\map P s$ is true.
That is, $s \notin S$.
This contradicts our assertion that $s \in S$.
Hence our assumption that $S \ne \O$ is false.
It follows by Proof by Contradiction that $S = \O$.
So $\map P n$ holds for all $n \in \N_{>0}$.
== Sources ==
* : Introduction $\S 4$: The natural numbers
* : $\S 2.3$: Theorem $2.18$
* : Chapter $3$: The Integers: $\S 9$. The Principles of Induction: Theorem $15$
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Theorem $4.2$
Category:Second Principle of Mathematical Induction
Second Principle of Mathematical Induction/Zero-Based

== Theorem ==
Let $\map P n$ be a propositional function depending on $n \in \N$.
Suppose that:
:$(1): \quad \map P 0$ is true
:$(2): \quad \forall k \in \N: \map P 0 \land \map P 1 \land \ldots \land \map P {k - 1} \land \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is true for all $n \in \N$.
== Proof ==
For each $n \in \N$, let $\map {P'} n$ be defined as:
:$\map {P'} n := \map P 0 \land \dots \land \map P n$
It suffices to show that $\map {P'} n$ is true for all $n \in \N$.
It is immediate from the assumption $\map P 0$ that $\map {P'} 0$ is true.
Now suppose that $\map {P'} n$ holds.
By $(2)$, this implies that $\map P {n + 1}$ holds as well.
Consequently, $\map {P'} n \land \map P {n + 1} = \map {P'} {n + 1}$ holds.
Thus by the Principle of Mathematical Induction:
:$\map {P'} n$ holds for all $n \in \N$
as desired.
== Also see ==
== Sources ==
* : $\S 1.2.1$: Mathematical Induction: Exercise $1$
Category:Second Principle of Mathematical Induction
Second Principle of Recursive Definition

== Theorem ==
Let $\N$ be the natural numbers.
Let $T$ be a set.
Let $a \in T$.
For each $n \in \N_{>0}$, let $G_n: T^n \to T$ be a mapping.
Then there exists exactly one mapping $f: \N \to T$ such that:
:$\forall x \in \N: \map f x = \begin{cases}
a & : x = 0 \\
\map {G_n} {\map f 0, \ldots, \map f n} & : x = n + 1
\end{cases}$
== Proof ==
Define $T^*$ to be the Kleene closure of $T$:
:$T^* := \ds \bigcup_{i \mathop = 1}^\infty T^i$
Note that, for convenience, the empty sequence is excluded from $T^*$.
Now define a mapping $\GG: T^* \to T^*$ by:
:$\map \GG {t_1, \ldots, t_n} = \tuple {t_1, \ldots, t_n, \map {G_n} {t_1, \ldots, t_n} }$
that is, extending each finite sequence $\tuple {t_1, \ldots, t_n}$ with the element $\map {G_n} {t_1, \ldots, t_n} \in T$.
By the Principle of Recursive Definition applied to $\GG$ and the finite sequence $\sequence a$, we obtain a unique mapping:
:$\FF: \N \to T^*: \map \FF x = \begin{cases} \sequence a & : x = 0 \\ \map \GG {\map \FF n} & : x = n + 1 \end {cases}$
Next define $f: \N \to T$ by:
:$\map f n = \text {the last element of $\map \FF n$}$
We claim that this $f$ has the sought properties, which will be proven by the Principle of Mathematical Induction.
We prove the following assertions by induction:
:$\map \FF n = \tuple {\map f 0, \map f 1, \ldots, \map f {n - 1}, \map {G_n} {\map f 0, \ldots, \map f {n - 1} } }$
:$\map f n = \map {G_n} {\map f 0, \ldots, \map f {n - 1} }$
For $n = 0$, these statements do not make sense, however it is immediate that $\map f 0 = \map {\operatorname {last} } {\sequence a} = a$.
For the base case, $n = 1$, we have:
Now assume that we have that:
:$\map \FF n = \tuple {\map f 0, \map f 1, \ldots, \map f {n - 1}, \map {G_n} {\map f 0, \ldots, \map f {n - 1} } }$
:$\map f n = \map {G_n} {\map f 0, \ldots, \map f {n - 1} }$
Then:
The result follows by the Principle of Mathematical Induction.
== Also known as ==
Some authors go through considerable effort to define the sequence $\sequence {G_n}_n$ as a single mapping $G$.
The domain of such a mapping is then for example given as one of the following:
:$\Dom G := \set {f: \N_{<n} \to T \mid n \in \N_{>0} } = \ds \bigcup_{n \mathop\in \N} \paren {\N_{<n} \to T}$
:$\Dom G := \ds \bigcup_{n \mathop\in \N_{>0} } T^n$
At we deem the presentation with separate $G_n$ to be more enlightening.
== Also see ==
* Principle of Recursive Definition
* Transfinite Recursion Theorem
== Sources ==
* : $1$: Set Theory and Logic: $\S 7$: Countable and Uncountable Sets
Category:Named Theorems
Category:Mapping Theory
Category:Natural Numbers
Second Principle of Transfinite Induction

== Theorem ==
Let $A$ be a class.
Let $\preccurlyeq$ be a well-ordering on $A$.
Let $P$ be a property that satisfies the following $3$ conditions:
:$(1): \quad P$ holds for the smallest element of $A$.
:$(2): \quad$ For all $x \in A$ which have an immediate successor $\map S x$, if $P$ holds for $x$, then $P$ holds for $\map S x$.
:$(3): \quad$ For all limit element $y$ of $A$, if $P$ holds for all $x \prec y$, then $P$ holds for $y$.
Then $P$ holds for all $x \in A$.
== Proof ==
 $P$ fails to hold for some $z \in A$.
The class of elements of $A$ for which $P$ does not hold is therefore non-empty.
We have that $A$ is a well-ordered class.
Hence there must be some smallest element $x \in A$ for which $P$ fails to hold.
By $(1)$, this cannot be the smallest element of $A$.
By $(2)$, this cannot be an immediate successor element.
By $(3)$, this cannot be a limit element.
From Categories of Elements under Well-Ordering, there are no other options for $x$.
Hence for all $x \in A$, , $P$ holds for $x$.
This contradicts our definition of $x$.
Hence there can be no such $z \in A$ for which $P$ fails to hold.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering: Proposition $1.9$ (Transfinite Induction Principle $2$)
Category:Well-Orderings
Category:Mathematical Induction
Second Projection on Ordered Pair of Sets

== Theorem ==
Let $a$ and $b$ be sets.
Let $w = \tuple {a, b}$ denote the ordered pair of $a$ and $b$.
Let $\map {\pr_2} w$ denote the second projection on $w$.
Then:
:$\ds \map {\pr_2} w = \begin {cases} \ds \map \bigcup {\bigcup w \setminus \bigcap w} & : \ds \bigcup w \ne \bigcap w \\ \ds \bigcup \bigcup w & : \bigcup w = \ds \bigcap w \end {cases}$
where:
:$\ds \bigcup$ and $\ds \bigcap$ denote union and intersection respectively.
:$\setminus$ denotes the set difference operator.
== Proof ==
We have by definition of second projection that:
:$\map {\pr_1} w = \map {\pr_1} {a, b} = b$
We consider:
Suppose $\ds \bigcup w \ne \bigcap w$.
Then:
demonstrating that the first case holds.
Now suppose that $\bigcup w = \bigcap w$.
Thus:
Hence:
The result follows;
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.6$: Functions: Exercise $1.6.6 \ \text{(ii)}$
Category:Ordered Pairs
Category:Projections
Second Ring Isomorphism Theorem/Also known as

== Second Ring Isomorphism Theorem: Also known as == 
The Second Ring Isomorphism Theorem is also referred to by some sources as the First Ring Isomorphism Theorem.
There is little consistency in the literature.
It can also be called the Second Isomorphism Theorem for Rings.
Because of the shape of its commutative diagram, this theorem is sometimes known as the Parallelogram Law.
However, that name is usually applied to a result in vector algebra, so use in this context is not recommended.
Category:Second Ring Isomorphism Theorem