hcju/proofwikiqa huggingface dataset:
Proof by induction: For all $n \in \N_{>0}$, let $\map P n$ be the proposition: :$\displaystyle \forall n \ge 1: \sum_{j \mathop = 1}^n j \paren {j + 1} = \frac {n \paren {n + 1} \paren {n + 2} } 3$ === Basis for the Induction === $\map P 1$ is true, as this just says $1 \times 2 = 2 = \dfrac {1 \times 2 \times 3} 3$. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \sum_{j \mathop = 1}^k j \paren {j + 1} = \frac {k \paren {k + 1} \paren {k + 2} } 3$ Then we need to show: :$\displaystyle \sum_{j \mathop = 1}^{k + 1} j \paren {j + 1} = \frac {\paren {k + 1} \paren {k + 2} \paren {k + 3} } 3$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^{k + 1} j \paren {j + 1}       | r = \sum_{j \mathop = 1}^k j \paren {j + 1} + \paren {k + 1} \paren {k + 2}       | c =  }} {{eqn | r = \frac {k \paren {k + 1} \paren {k + 2} } 3 + \paren {k + 1} \paren {k + 2}       | c = Induction hypothesis }} {{eqn | r = \frac {k \paren {k + 1} \paren {k + 2} + 3 \paren {k + 1} \paren {k + 2} } 3       | c =  }} {{eqn | r = \frac {\paren {k + 1} \paren {k + 2} \paren {k + 3} } 3       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \ge 1: \sum_{j \mathop = 1}^n j \paren {j + 1} = \frac {n \paren {n + 1} \paren {n + 2} } 3$ {{qed}}
Observe that: {{begin-eqn}} {{eqn | l = 3 i \left({i + 1}\right)       | r = i \left({i + 1}\right) \left({i + 2}\right) - i \left({i + 1}\right) \left({i - 1}\right) }} {{eqn | r = \left({i + 1}\right) \left({\left({i + 1}\right) + 1}\right) \left({\left({i + 1}\right) - 1}\right) - i \left({i + 1}\right) \left({i - 1}\right) }} {{end-eqn}} Thus: :$3 i \left({i + 1}\right) = b \left({i + 1}\right) - b \left({i}\right)$ where: :$b \left({i}\right) = i \left({i + 1}\right) \left({i - 1}\right)$ This can therefore be used as the basis of a telescoping series, as follows: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n 3 i \left({i + 1}\right)       | r = \sum_{i \mathop = 1}^n \left({i \left({i + 1}\right) \left({i + 2}\right) - i \left({i + 1}\right) \left({i - 1}\right)}\right) }} {{eqn | r = n \left({n + 1}\right) \left({n + 2}\right) - 1 \left({1 + 1}\right) \left({1 - 1}\right)       | c = Telescoping Series }} {{eqn | r = n \left({n + 1}\right) \left({n + 2}\right) }} {{end-eqn}} Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^n j \left({j + 1}\right)       | r = \sum_{j \mathop = 1}^n \left({j^2 + j}\right)       | c =  }} {{eqn | r = \sum_{j \mathop = 1}^n j^2 + \sum_{j \mathop = 1}^n j       | c =  }} {{eqn | r = \sum_{j \mathop = 1}^n j^2 + \frac {n \left({n + 1}\right)} 2       | c = Closed Form for Triangular Numbers }} {{eqn | r = \frac {n \left({n + 1}\right) \left({2 n + 1}\right)} 6 + \frac {n \left({n + 1}\right)} 2       | c = Sum of Sequence of Squares }} {{eqn | r = n \left({n + 1}\right) \left({\frac {\left({2 n + 1}\right)} 6 + \frac 1 2}\right)       | c = Extracting $n \left({n + 1}\right)$ as a common factor }} {{eqn | r = \frac {n \left({n + 1}\right) \left({n + 2}\right)} 3       | c = Simplification }} {{end-eqn}}
Proof by induction: For all $n \in \N_{>0}$, let $\map P n$ be the proposition: :$\displaystyle \forall n \ge 1: \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} } = \frac n {n + 1}$ === Basis for the Induction === $\map P 1$ is true, as this just says $\dfrac 1 2 = \dfrac 1 2$. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \sum_{j \mathop = 1}^k \frac 1 {j \paren {j + 1} } = \frac k {k + 1}$ Then we need to show: :$\displaystyle \sum_{j \mathop = 1}^{k + 1} \frac 1 {j \paren {j + 1} } = \frac {k + 1} {k + 2}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^{k + 1} \frac 1 {j \left({j + 1}\right)}       | r = \sum_{j \mathop = 1}^k \frac 1 {j \paren {j + 1} } + \frac 1 {\paren {k + 1} \paren {k + 2} }       | c =  }} {{eqn | r = \frac k {k + 1} + \frac 1 {\paren {k + 1} \paren {k + 2} }       | c = Induction hypothesis }} {{eqn | r = \frac {k \paren {k + 2} + 1} {\paren {k + 1} \paren {k + 2} }       | c =  }} {{eqn | r = \frac {k^2 + 2 k + 1} {\paren {k + 1} \paren {k + 2} }       | c =  }} {{eqn | r = \frac {\paren {k + 1}^2} {\paren {k + 1} \paren {k + 2} }       | c =  }} {{eqn | r = \frac {k + 1} {k + 2}       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \ge 1: \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} } = \frac n {n + 1}$ {{qed}}
We can observe that: :$\dfrac 1 {j \paren {j + 1} } = \dfrac 1 j - \dfrac 1 {j + 1}$ and that $\displaystyle \sum_{j \mathop = 1}^n \paren {\frac 1 j - \frac 1 {j + 1} }$ is a telescoping series.  Therefore: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} }       | r = \sum_{j \mathop = 1}^n \paren {\frac 1 j - \frac 1 {j + 1} }       | c = Telescoping Series: Example 1 }} {{eqn | r = 1 - \frac 1 {n + 1}       | c =  }} {{eqn | r = \frac n {n + 1}       | c =  }} {{end-eqn}} {{qed}}
Proof by induction: For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\displaystyle \sum_{j \mathop = 0}^n F_j = F_{n + 2} - 1$ $\map P 0$ is the case: {{begin-eqn}} {{eqn | l = F_0       | r = 0       | c =  }} {{eqn | r = 1 - 1       | c =  }} {{eqn | r = F_2 - 1       | c =  }} {{end-eqn}} which is seen to hold. === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = F_1       | r = 1       | c =  }} {{eqn | r = 2 - 1       | c =  }} {{eqn | r = F_3 - 1       | c =  }} {{end-eqn}} which is seen to hold. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \sum_{j \mathop = 1}^k F_j = F_{k + 2} - 1$ Then we need to show: :$\displaystyle \sum_{j \mathop = 1}^{k + 1} F_j = F_{k + 3} - 1$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^{k + 1} F_j       | r = \sum_{j \mathop = 1}^k F_j + F_{k + 1}       | c =  }} {{eqn | r = F_{k + 2} - 1 + F_{k + 1}       | c = Induction Hypothesis }} {{eqn | r = F_{k + 3} - 1       | c = {{Defof|Fibonacci Number}} }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 0}: \sum_{j \mathop = 0}^n F_j = F_{n + 2} - 1$ {{qed}}
Proof by induction: For all $n \in \N_{>0}$, let $\map P n$ be the proposition: :$\displaystyle \sum_{j \mathop = 1}^n F_{2 j - 1} = F_{2 n}$ === Basis for the Induction === $\map P 1$ is the case $F_1 = 1 = F_2$, which holds from the definition of Fibonacci numbers. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \sum_{j \mathop = 1}^k F_{2 j - 1} = F_{2 k}$ Then we need to show: :$\displaystyle \sum_{j \mathop = 1}^{k + 1} F_{2 j - 1} = F_{2 k + 2}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^{k + 1} F_{2 j - 1}       | r = \sum_{j \mathop = 1}^k F_{2 j - 1} + F_{2 k + 1}       | c =  }} {{eqn | r = F_{2 k} + F_{2 k + 1}       | c = Induction Hypothesis }} {{eqn | r = F_{2 k + 2}       | c = Definition of Fibonacci Numbers }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \ge 1: \sum_{j \mathop = 1}^n F_{2 j - 1} = F_{2 n}$ {{qed}}
Proof by induction: For all $n \in \N_{>0}$, let $\map P n$ be the proposition: :$\displaystyle \sum_{j \mathop = 1}^n F_{2 j} = F_{2 n + 1} - 1$ === Basis for the Induction === $\map P 1$ is the case $F_2 = 1 = F_3 - 1$, which holds from the definition of Fibonacci numbers. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \sum_{j \mathop = 1}^k F_{2 j} = F_{2 k + 1} - 1$ Then we need to show: :$\displaystyle \sum_{j \mathop = 1}^{k + 1} F_{2 j} = F_{2 k + 3} - 1$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^{k + 1} F_{2 j}       | r = \sum_{j \mathop = 1}^k F_{2 j} + F_{2 \paren {k + 1} }       | c =  }} {{eqn | r = F_{2 k + 1} - 1 + F_{2 k + 2}       | c = Induction Hypothesis }} {{eqn | r = F_{2 k + 3} - 1       | c = {{Defof|Fibonacci Number}} }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: : $\displaystyle \forall n \ge 1: \sum_{j \mathop = 1}^n F_{2 j} = F_{2 n + 1} - 1$ {{qed}}
Proof by induction: For all $n \in \N_{>0}$, let $\map P n$ be the proposition: :$\ds \sum_{j \mathop = 1}^{2 n - 1} F_j F_{j + 1} = {F_{2 n} }^2$ === Basis for the Induction === $\map P 1$ is true, as this just says $F_1 F_2 = 1 \times 1 = 1 = {F_2}^2$. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\ds \sum_{j \mathop = 1}^{2 k - 1} F_j F_{j + 1} = {F_{2 k} }^2$ Then we need to show: :$\ds \sum_{j \mathop = 1}^{2 k + 1} F_j F_{j + 1} = {F_{2 \paren {k + 1} } }^2$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^{2 k + 1} F_j F_{j + 1}       | r = \sum_{j \mathop = 1}^{2 k - 1} F_j F_{j + 1} + F_{2 k} F_{2 k + 1} + F_{2 k + 1} F_{2 k + 2}       | c =  }} {{eqn | r = {F_{2 k} }^2 + F_{2 k} F_{2 k + 1} + F_{2 k + 1} F_{2 k + 2}       | c = Induction hypothesis }} {{eqn | r = F_{2 k} \paren {F_{2 k} + F_{2 k + 1} } + F_{2 k + 1} F_{2 k + 2}       | c =  }} {{eqn | r = F_{2 k} F_{2 k + 2} + F_{2 k + 1} F_{2 k + 2}       | c =  }} {{eqn | r = F_{2 k + 2} \paren {F_{2 k} + F_{2 k + 1} }       | c =  }} {{eqn | r = {F_{2 k + 2} }^2       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\ds \forall n \ge 1: \sum_{j \mathop = 1}^{2 n - 1} F_j F_{j + 1} = {F_{2 n} }^2$ {{qed}}
Proof by induction: For all $n \in \N_{>0}$, let $\map P n$ be the proposition: :$L_n = F_{n - 1} + F_{n + 1}$ === Basis for the Induction === $\map P 1$ is true, as this just says: :$L_1 = 1 = F_0 + F_2$ which holds by definition of the Fibonacci numbers. This is our basis for the induction. === Induction Hypothesis === Let us make the supposition that, for some $k \in \N: k \ge 1$, the proposition $\map P j$ holds for all $j \in \N: 1 \le j \le k$. We shall show that it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$L_j = F_{j - 1} + F_{j + 1}$ Then we need to show: :$L_{k + 1} = F_k + F_{k + 2}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = L_{k + 1}       | r = L_{k - 1} + L_k       | c =  }} {{eqn | r = F_{k - 2} + F_k + F_{k - 1} + F_{k + 1}       | c = Induction Hypothesis }} {{eqn | r = \paren {F_{k - 2} + F_{k - 1} } + \paren {F_k + F_{k + 1} }       | c =  }} {{eqn | r = F_k + F_{k + 2}       | c =  }} {{end-eqn}} So $\map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction. Therefore: : $\forall n \in \N, n \ge 1: L_n = F_{n - 1} + F_{n + 1}$ {{Qed}} Category:Lucas Numbers Category:Fibonacci Numbers kyz461ovjq23d9urgofm3aen8a8pjxz
Proof by induction: For all $\forall n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \sum_{j \mathop = 1}^n 2^{j - 1} L_j = 2^n F_{n + 1} - 1$ === Basis for the Induction === $P(1)$ is true, as this just says $L_1 = 1 = 2^1 \times F_1 - 1$ which follows directly from the definitions of Fibonacci numbers and Lucas numbers. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$\displaystyle \sum_{j \mathop = 1}^k 2^{j - 1} L_j = 2^k F_{k + 1} - 1$ Then we need to show: :$\displaystyle \sum_{j \mathop = 1}^{k + 1} 2^{j - 1} L_j = 2^{k + 1} F_{k + 2} - 1$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^{k + 1} 2^{j - 1} L_j       | r = \sum_{j \mathop = 1}^k 2^{j - 1} L_j + 2^k L_{k + 1}       | c =  }} {{eqn | r = 2^k F_{k + 1} - 1 + 2^k L_{k + 1}       | c = Induction hypothesis }} {{eqn | r = 2^k \paren {F_{k + 1} + L_{k + 1} } - 1       | c =  }} {{eqn | r = 2^k \paren {F_{k + 1} + F_k + F_{k + 2} } - 1       | c = Lucas Number as Sum of Fibonacci Numbers }} {{eqn | r = 2^k \paren {F_{k + 2} + F_{k + 2} } - 1       | c =  }} {{eqn | r = 2^{k + 1} F_{k + 2} - 1       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: : $\displaystyle \forall n \in \N_{>0}: \sum_{j \mathop = 1}^n 2^{j - 1} L_j = 2^n F_{n + 1} - 1$ {{qed}}
Given that $n \in \N$ be odd, it can be expressed in the form: :$n = 2 m + 1$ where $m \in \N$. The proof proceeds by strong induction. For all $m \in \N$, let $\map P m$ be the proposition: : $x^{2 m + 1} + y^{2 m + 1} = \paren {x + y} \paren {x^{2 m} + \cdots + y^{2 m} }$ $\map P 0$ is the case: : $x + y = x + y$ which is trivially an identity. === Basis for the Induction === $\map P 1$ is the case: :$x^3 + y^3 = \paren {x + y} \paren {x^2 - x y + y^2}$ which follows by Sum of Two Cubes. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\exists k \in \N: \forall j: 1 \le j \le k: x^{2 j + 1} + y^{2 j + 1} = \paren {x + y} \map {P_{2 j} } {x, y}$ where $\map {P_{2 j} } {x, y}$ is a polynomial of degree $2 j$ in $x$ and $y$. from which it is to be shown that: :$x^{2 k + 3} + y^{2 k + 3} = \paren {x + y} \map {P_{2 k + 2} } {x, y}$ where $\map {P_{2 k + 2} } {x, y}$ is a polynomial of degree $2 k + 2$ in $x$ and $y$. === Induction Step === This is the induction step: We have that: :$\paren {x^{2 k + 1} + y^{2 k + 1} } \paren {x^2 + y^2} = x^{2 k + 3} + y^{2 k + 3} + x^2 y^{2 k + 1} + x^{2 k + 1} y^2$ So: {{begin-eqn}} {{eqn | l = x^{2 k + 3} + y^{2 k + 3}       | r = \paren {x + y} \paren {x^2 + y^2} \map {P_{2 k} } {x, y} - x^2 y^{2 k + 1} - x^{2 k + 1} y^2       | c =  }} {{eqn | r = \paren {x + y} \paren {x^2 + y^2} \map {P_{2 k} } {x, y} - x^2 y^2 \paren {x^{2 k - 1} + y^{2 k - 1} }       | c =  }} {{end-eqn}} But $\paren {x^{2 k - 1} + y^{2 k - 1} }$ itself is of the form: :$\paren {x + y} \map {P_{2 k - 2} } {x, y}$ So: :$x^{2 k + 3} + y^{2 k + 3} = \paren {x + y} \paren {\paren {x^2 + y^2} \map {P_{2 k} } {x, y} - x^2 y^2 \map {P_{2 k - 2} } {x, y} }$ Hence the result by the Second Principle of Mathematical Induction. {{qed}}
=== Reverse Implication === {{:Condition for Well-Foundedness/Reverse Implication}} === Forward Implication === {{:Condition for Well-Foundedness/Forward Implication}} {{ADC|Condition for Well-Foundedness/Reverse Implication}}
From Subset Relation is Ordering, we have that $\subseteq$ is an ordering on any set of subsets of a given set. Suppose $S$ is neither a singleton nor the empty set. Then $\exists a, b \in S$ such that $a \ne b$. Then $\set a \in \powerset S$ and $\set b \in \powerset S$. However, $\set a \nsubseteq \set b$ and $\set b \nsubseteq \set a$. So by definition, $\subseteq$ is a partial ordering. Now suppose $S = \O$. Then $\powerset S = \set \O$ and, by Empty Set is Subset of All Sets, $\O \subseteq \O$. Hence, trivially, $\subseteq$ is a total ordering on $\powerset S$. Now suppose $S$ is a singleton: let $S = \set a$. Then $\powerset S = \set {\O, \set a}$. So there are only two elements of $\powerset S$, and we see that $\O \subseteq \set a$ from Empty Set is Subset of All Sets. So, trivially again, $\subseteq$ is a total ordering on $\powerset S$. {{qed}}
From Subset Relation is Ordering, we have that $\struct {\mathbb S, \subseteq}$ is an ordered set. For each $a \in S$, let $a^\prec$ be the lower closure of $a$: :$a^\prec := \set {b \in S: b \preceq a}$ Then let $T$ be defined as: :$T := \set {a^\prec: a \in S}$ Let the mapping $\phi: S \to T$ be defined as: :$\map \phi a = a^\prec$ We are to show that $\phi$ is an order isomorphism. $\phi$ is clearly surjective, as every $a^\prec$ is defined from some $a \in S$. Now suppose $x^\prec, y^\prec \in T: x^\prec = y^\prec$. Then: :$\set {b \in S: b \preceq x} = \set {b \in S: b \preceq y}$ We have that: :$x \in x^\prec = y^\prec$ and $y \in y^\prec = x^\prec$ which means: :$x \preceq y$ and $y \preceq x$ So as an ordering is antisymmetric, we have $x = y$ and so $\phi$ is injective. Hence by definition, $\phi$ is a bijection. Now let $a_1 \preceq a_2$. Then by definition: :$a_1 \in {a_2}^\prec$ Let $a_3 \in {a_1}^\prec$. Then by definition: :$a_3 \preceq a_1$ As an ordering is transitive, it follows that: :$a_3 \preceq a_2$ and so: :$a_3 \in {a_2}^\prec$ So by definition of a subset: :${a_1}^\prec \subseteq {a_2}^\prec$ Therefore, $\phi$ is order-preserving Conversely, suppose that ${a_1}^\prec \subseteq {a_2}^\prec$. Then, since $a_1 \in {a_1}^\prec$, also $a_1 \in {a_2}^\prec$ by definition of subset. By definition of ${a_2}^\prec$: :$a_1 \preceq a_2$ Hence it is seen that $\phi^{-1}$ is also order-preserving. Thus it follows that $\phi$ is an order isomorphism between $\struct {S, \preceq}$ and $\struct {\mathbb S, \subseteq}$. {{qed}}
From Subset Relation is Ordering, we have that $\struct {\mathbb S, \subseteq}$ is an ordered set. We are to show that $\phi$ is an order isomorphism. $\phi$ is clearly surjective, as every $a^\preceq$ is defined from some $a \in S$. By the Lemma, $\phi$ is order-preserving. Suppose that ${a_1}^\preceq \subseteq {a_2}^\preceq$. We have that: :$a_1 \in {a_1}^\preceq$ Thus by definition of subset: :$a_1 \in {a_2}^\preceq$ By definition of ${a_2}^\preceq$: :$a_1 \preceq a_2$ Thus $\phi$ is also order-reflecting. Thus it follows that $\phi$ is an order isomorphism between $\struct {S, \preceq}$ and $\struct {\mathbb S, \subseteq}$. {{qed}}
To establish that $\subseteq$ is an ordering, we need to show that it is reflexive, antisymmetric and transitive. So, checking in turn each of the criteria for an ordering: === Reflexivity === {{begin-eqn}} {{eqn | lo= \forall T \in \mathbb S:       | l = T       | o = \subseteq       | r = T       | c = Set is Subset of Itself }} {{end-eqn}} So $\subseteq$ is reflexive. {{qed|lemma}} === Antisymmetry === {{begin-eqn}} {{eqn | lo= \forall S_1, S_2 \in \mathbb S:       | l = S_1 \subseteq S_2 \land S_2 \subseteq S_1       | o = \iff       | r = S_1 = S_2       | c = {{Defof|Set Equality|index = 2}} }} {{end-eqn}} So $\subseteq$ is antisymmetric. {{qed|lemma}} === Transitivity === {{begin-eqn}} {{eqn | lo= \forall S_1, S_2, S_3 \in \mathbb S:       | l = S_1 \subseteq S_2 \land S_2 \subseteq S_3       | o = \implies       | r = S_1 \subseteq S_3       | c = Subset Relation is Transitive }} {{end-eqn}} That is, $\subseteq$ is transitive. {{qed|lemma}} So we have shown that $\subseteq$ is an ordering on $\mathbb S$. {{qed}}
{{AimForCont}} that $T \ne S$. From Set Difference is Subset, $S \setminus T \subset S$. From Set Difference with Proper Subset, $S \setminus T \ne \O$. By the definition of a well-ordered set, there exists a smallest element $s$ of $S \setminus T$. As $s \in S$, it follows from the definition of $T$ that: :$\forall t \in S: t \prec s \implies t \in T$ But then $s \in T$ by hypothesis, contradicting the definition of $s$. Hence the result, by Proof by Contradiction. {{qed}}
Let $T = \set {x \in S: \map f x \prec x}$. We are to show that $T = \O$. {{AimForCont}} that $T \ne \O$. Then as $\struct {S, \preceq}$ is a woset, by definition $T$ has a minimal element: call it $x_0$. Since $x_0 \in T$, we have $\map f {x_0} \prec x_0$. So, let $x_1 = \map f {x_0}$. $f$ is an order isomorphism, so since $x_1 \prec x_0$, it follows that $\map f {x_1} \prec \map f {x_0} = x_1$. So as $\map f {x_1} \prec x_1$ it follows that $x_1 \in T$. But $x_0$ was chosen to be the minimal element of $T$. From this contradiction, it follows that it can not be the case that $T \ne \O$. So $T = \O$ and so $\forall x \in S: x \preceq \map f x$. {{qed}}
Let $f: S_1 \to S_2$ and $g: S_1 \to S_2$ both be order isomorphisms. By Inverse of Order Isomorphism is Order Isomorphism, the inverse $f^{-1}$ is also an order isomorphism. Let $h = f^{-1} \circ g$ be the composition of $f^{-1}$ and $g$, which, by Composite of Order Isomorphisms is Order Isomorphism, is itself an order isomorphism. So, by Order Isomorphism from Woset onto Subset: :$\forall x \in S_1: x \preceq_1 \map h x$ Now we apply $f$ to $S_1$ and see that: :$\forall x \in S_1: \map f x \preceq_2 \map f {\map h x} = \map g x$ In a similar way we can show that: :$\forall x \in S_1: \map g x \preceq_2 \map f x$ Hence $f = g$ and the proof is complete. {{qed}}
{{AimForCont}} $f: S \to S_a$ is an order isomorphism. By Order Isomorphism from Woset onto Subset, $\forall x \in S: x \preceq \map f x$. In particular, then, $a \preceq \map f a$. But the codomain of $f$ is $S_a$, so $\map f a \in S_a$. Thus $\map f a \prec a$, which is a contradiction. So there can be no such order isomorphism. {{qed}}
Define $f: S \to A$ as: :$\forall a \in S: \map f a = a^\prec$ where $a^\prec$ is the initial segment determined by $a$. === $f$ is Surjective === $f$ is trivially surjective by the definition of $A$. {{Qed|lemma}} === $f$ is Strictly Increasing === Let $x, y \in S$ with $x \prec y$. Let $z \in \map f x$. Then by the definition of initial segment: :$z \prec x$ By Reflexive Reduction of Ordering is Strict Ordering, $\prec$ is also transitive. Thus: :$z \prec y$ Thus by the definition of initial segment: :$z \in y^\prec = \map f y$ As this holds for all such $z$: :$\map f x \subseteq \map f y$ As $x \prec y$: :$x \in y^\prec = \map f y$ But since $\prec$ is antireflexive: :$x \nprec x$ so: :$x \notin \map f x$ Thus: :$\map f x \subsetneqq \map f y$ As this holds for all such $x$ and $y$, $f$ is strictly increasing. {{qed|lemma}} Since a well-ordering is a total ordering, $f$ is an order embedding by Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing. Thus $f$ is a surjective order embedding and therefore an order isomorphism. {{qed}}
Let $S$ be an ordinal. Let the minimal element of $S$ be $s_0$. This exists by dint of an ordinal being a woset. From Ordering on Ordinal is Subset Relation, $S$ is well-ordered by $\subseteq$. So, by definition of an ordinal: :$s_0 = \set {s \in S: s \subset s_0}$ But as $s_0$ is minimal, there ''are'' no elements of $S$ which are a subset of it. So: :$\set {s \in S: s \subset s_0} = \O$ Hence the result. {{qed}}
The first equivalence is an immediate consequence of Equivalence of Definitions of Ordinal. The second equivalence holds for any well-ordered set by Woset Isomorphic to Set of its Sections. The third equivalence holds by definition of an ordinal. It follows from Ordering is Equivalent to Subset Relation and Order Isomorphism between Wosets is Unique that this ordering is the only one. {{qed}}