hcju/proofwikiqa huggingface dataset:
Let: :$\mathbf a = \begin{bmatrix} a_x \\ a_y \\a_z \end{bmatrix}$, $\mathbf b = \begin{bmatrix} b_x \\ b_y \\ b_z \end{bmatrix}$, $\mathbf c = \begin{bmatrix} c_x \\ c_y \\ c_z \end{bmatrix}$ be vectors in $\R^3$. Then: {{begin-eqn}} {{eqn | l = \mathbf a \times \paren {\mathbf b + \mathbf c}       | r = \begin{bmatrix} a_x \\ a_y \\a_z \end{bmatrix} \times \paren {\begin{bmatrix} b_x \\ b_y \\ b_z \end{bmatrix} + \begin{bmatrix} c_x \\ c_y \\ c_z \end{bmatrix} } }} {{eqn | r = \begin{bmatrix} a_x \\ a_y \\a_z \end{bmatrix} \times {\begin{bmatrix} b_x + c_x \\ b_y + c_y \\ b_z + c_z \end{bmatrix} }       | c = {{Defof|Vector Sum}} }} {{eqn | r = \begin{bmatrix} a_y \paren {b_z + c_z} - a_z \paren {b_y + c_y} \\ a_z \paren {b_x + c_x} - a_x \paren {b_z + c_z} \\ a_x \paren {b_y + c_y} - a_y \paren {b_x + c_x} \end{bmatrix}       | c = {{Defof|Vector Cross Product}} }} {{eqn | r = \begin{bmatrix} a_y b_z + a_y c_z - a_z b_y - a_z c_y \\ a_z b_x + a_z c_x - a_x b_z - a_x c_z \\ a_x b_y + a_x c_y - a_y b_x - a_y c_x \end{bmatrix}       | c = Real Multiplication Distributes over Addition }} {{eqn | r = \begin{bmatrix} a_y b_z - a_z b_y + a_y c_z - a_z c_y \\ a_z b_x - a_x b_z + a_z c_x - a_x c_z \\ a_x b_y - a_y b_x + a_x c_y - a_y c_x \end{bmatrix}       | c = Real Addition is Commutative }} {{eqn | r = \begin{bmatrix} a_y b_z - a_z b_y \\ a_z b_x - a_x b_z \\ a_x b_y - a_y b_x \end{bmatrix} + \begin{bmatrix} a_y c_z - a_z c_y \\ a_z c_x - a_x c_z \\ a_x c_y - a_y c_x \end{bmatrix}       | c = {{Defof|Vector Sum}} }} {{eqn | r = \paren {\begin{bmatrix}a_x \\ a_y \\ a_z \end{bmatrix} \times \begin{bmatrix} b_x \\ b_y \\ b_z \end{bmatrix} } + \paren {\begin{bmatrix} a_x \\ a_y \\ a_z \end{bmatrix} \times \begin{bmatrix} c_x \\ c_y \\ c_z \end{bmatrix} }       | c = {{Defof|Vector Cross Product}} }} {{eqn|r = \paren {\mathbf a \times \mathbf b} + \paren {\mathbf a \times \mathbf c} }} {{end-eqn}} {{qed}}
Let Euler's equation be :$\map {F_y} {x, \hat y, \hat y'} - \dfrac \d {\d x} \map {F_{y'} } {x, \hat y, \hat y'} = 0$ which is derived from: :$\displaystyle \int_a^b \paren {\map {F_y} {x, \hat y, \hat y'} - \frac \d {\d x} \map {F_{y'} } {x, \hat y, \hat y'} } \rd x = 0$ Let $\map {\hat y} x = \map y x$ and $\map {\hat y} x = \map y x + \map h x$ be solutions of Euler's equation. By Taylor's Theorem: {{begin-eqn}} {{eqn | r = \map {F_y} {x, y + h, y' + h'} - \frac \d {\d x} \map {F_{y'} } {x, y + h, y' + h'}       | o =  }} {{eqn | r = F_y + F_{yy} h + F_{yy'} h' + \map \OO {h^2, h'^2} - \map {\frac \d {\d x} } {F_{y'} + F_{y'y} h + F_{y'y'} h' + \map \OO {h^2, h'^2} } }} {{eqn | r = \paren {F_y - \frac \d {\d x} F_{y'} } + F_{yy} h + \map \OO {h^2, h h', h'^2} - F_{y'y}' h - F_{y'y'}' h' - F_{y'y'} h'' - \frac \d {\d x} \map \OO {h^2, h h', h'^2} }} {{eqn | r = \paren {F_{yy} - \frac \d {\d x} F_{y'y} } h - \map {\frac \d {\d x} } {F_{y'y'} h'} + \map \OO {h^2, h h', h'^2} }} {{end-eqn}} where the omitted ordered tuple of variables is $\tuple {x, y, y'}$, and $\map {\hat y} x = \map y x$ has been used as a solution to $F_y - \dfrac \d {\d x} F_{y'} = 0$. Therefore, Euler's equation is to be derived from :$\displaystyle \int_a^b \paren {\paren {F_{yy} - \frac \d {\d x} F_{y'y} } h - \map {\frac \d {\d x} } {F_{y'y'} h'} + \map \OO {h^2, h h', h'^2} } \rd x = 0$ By integration by parts,  :$\displaystyle \int_a^b \map \OO {h^2, h h', h'^2} \rd x = \int_a^b \map \OO {h^2} \rd x$ Thus, the equivalent differential equation is: :$\paren {F_{yy} - \dfrac \d {\d x} F_{y'y} } h - \map {\dfrac \d {\d x} } {F_{y'y'} h'} + \map \OO {h^2} = 0$ Omission of $\map \OO {h^2}$ and multiplication of equation by $\frac 1 2$ yields Jacobi's equation. {{qed}}
By Bottom is Compact: :$\bot$ is a compact element. By definition of the smallest element: :$\bot \preceq x$ Thus by definition of compact closure: :$\bot \in x^{\mathrm{compact} }$ {{qed}} Category:Way Below Relation mbp6qkcdgmumov8gk4ujkel15fpykgp
Taking each triplet in turn: {{begin-eqn}} {{eqn | l = 33       | r = 3 \times 11 }} {{eqn | l = 34       | r = 2 \times 17 }} {{eqn | l = 35       | r = 5 \times 7 }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 85       | r = 5 \times 17 }} {{eqn | l = 86       | r = 2 \times 43 }} {{eqn | l = 87       | r = 3 \times 29 }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 93       | r = 3 \times 31 }} {{eqn | l = 94       | r = 2 \times 47 }} {{eqn | l = 95       | r = 5 \times 19 }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 141       | r = 3 \times 47 }} {{eqn | l = 142       | r = 2 \times 71 }} {{eqn | l = 143       | r = 11 \times 13 }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 201       | r = 3 \times 67 }} {{eqn | l = 202       | r = 2 \times 101 }} {{eqn | l = 203       | r = 7 \times 29 }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 213       | r = 3 \times 71 }} {{eqn | l = 214       | r = 2 \times 107 }} {{eqn | l = 215       | r = 5 \times 43 }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 217       | r = 7 \times 31 }} {{eqn | l = 218       | r = 2 \times 109 }} {{eqn | l = 219       | r = 3 \times 73 }} {{end-eqn}} It is noted that the triplet: :$121, 122, 123$ while consisting of semiprimes, not all of these are the product of $2$ distinct prime numbers, as $121 = 11^2$. {{qed}}
Let $n = p \times q$ where $p$ and $q$ are primes. We have by definition of divisor: {{begin-eqn}} {{eqn | l = 1       | o = \divides       | r = n       | c = One Divides all Integers }} {{eqn | l = p       | o = \divides       | r = n       | c = {{Defof|Divisor of Integer}} }} {{eqn | l = q       | o = \divides       | r = n       | c = {{Defof|Divisor of Integer}} }} {{eqn | l = p \times q       | o = \divides       | r = n       | c = Integer Divides Itself }} {{end-eqn}} where $\divides$ denotes divisibility. Suppose $a \divides n$ such that $1 \le a < n$. Suppose $a \ne p$. By definition of prime number: :$a \perp p$ where $\perp$ denotes coprimality. From Euclid's Lemma: :$a \divides q$ and so by definition of prime number: :$a = q$ or: :$a = 1$ Similarly, suppose $a \ne q$. By the same reasoning: :$a = p$ or: :$a = 1$ Thus the only positive divisors are as above. {{qed}} {{questionable|The above is of course rubbish.}}
Let $n = p \times q$ where $p$ and $q$ are primes. From Product of Two Distinct Primes has 4 Positive Divisors, the positive divisors of $n$ are: :$1, p, q, pq$ Thus the product of all the divisors of $n$ is: :$1 \times p \times q \times p q = p^2 q^2 = n^2$ Hence the result, by definition of multiplicatively perfect. {{qed}}
Let $n = p^3$ where $p$ is prime. The positive divisors of $n$ are: :$1, p, p^2, p^3$ This result follows from Divisors of Power of Prime. {{qed}}
Let $n = p^3$ where $p$ is prime. From Cube of Prime has 4 Positive Divisors, the positive divisors of $n$ are: :$1, p, p^2, p^3$ Thus the product of all the divisors of $n$ is: :$1 \times p \times p^2 \times p^3 = p^6 = n^2$ Hence the result, by definition of multiplicatively perfect. {{qed}}
By Bottom in Compact Closure: :$\bot \in x^{\mathrm{compact} }$ where $\bot$ denotes the smallest element in $L$. Thus by definition: :$x^{\mathrm{compact} }$ is non-empty. Let $y, z \in x^{\mathrm{compact} }$ By definition of compact closure: :$y$ and $z$ are compact elements and $y \preceq x$ and $z \preceq x$ By definitions of supremum and upper bound: :$y \vee z \preceq x$ By definition of compact element: :$y \ll y$ and $z \ll z$ where $\ll$ denotes the way below relation. By Join Succeeds Operands: :$y \preceq y \vee z$ and $z \preceq y \vee z$ By Preceding and Way Below implies Way Below and definition of reflexivity: :$y \ll y \vee z$ and $z \ll y \vee z$ By Join is Way Below if Operands are Way Below: :$y \vee z \ll y \vee z$ By definition: :$y \vee z$ is compact element. By definition of compact closure: :$y \vee z \in x^{\mathrm{compact} }$ Thus :$\exists v \in x^{\mathrm{compact} }: y \preceq v \land z \preceq v$ {{qed}}
The sequence of lucky numbers begins: :$1, 3, 7, 9, 13, 15, 21, 25, 31, 33, \ldots$ Thus we have: {{begin-eqn}} {{eqn | l = 34       | r = 1 + 33       | c =  }} {{eqn | r = 3 + 31       | c =  }} {{eqn | r = 9 + 25       | c =  }} {{eqn | r = 13 + 21       | c =  }} {{end-eqn}} {{qed}}
It follows from Integers form Ordered Integral Domain that for any integer $m$ either: :$m = 0$ or :$m > 0$ or: :$m < 0$ === Case 1 === By Element to the Power of Zero is Identity: : $m = 0 \implies \sigma^m = e$ So: :$\sigma^m \paren i = e \paren i = i$ {{qed|lemma}} === Case 2 === Follows from Fixed Point of Mappings is Fixed Point of Composition: General Result. {{qed|lemma}} === Case 3 === For all $m < 0$, $m = -k$ for some (strictly) positive integer $k$. Therefore: :$\sigma^k \paren i = i \implies i = \sigma^{-k} \paren i$ Hence the result. {{qed}} Category:Symmetric Groups Category:Permutation Theory 4ahdh5adekok0tt89pzs0c62sqq7u89
Let $\mathbf A$ be a real symmetric matrix. Then we have: {{begin-eqn}} {{eqn | l = \left[{\mathbf A}\right]^\dagger_{ij}       | r = \overline{\left[{\mathbf A}\right]_{ji} }       | c = {{Defof|Hermitian Conjugate}} }} {{eqn | r = \left[{\mathbf A}\right]_{ji}       | c = $\mathbf A$ is real }} {{eqn | r = \left[{\mathbf A}\right]_{ij}       | c = $\mathbf A$ is symmetric }} {{end-eqn}} So: :$\mathbf A = \mathbf A^\dagger$ Thus, by definition, $\mathbf A$ is Hermitian. {{qed}} Category:Linear Algebra Category:Hermitian Matrices gxilvi28qzxafxfs9ovlunu7gvs8ees
We will prove that :$f$ is an order embedding. Let $x, y \in S$. '''Sufficient condition''' Assume that :$x \preceq y$ By Compact Closure is Increasing: :$x^{\mathrm {compact} } \subseteq y^{\mathrm {compact} }$ By definition of $f$: :$\map f x \subseteq \map f y$ Thus by definition of inclusion ordered set: :$\map f x \precsim \map f y$ {{qed|lemma}} '''Necessary condition''' Assume that: :$\map f x \precsim \map f y$ By definition of inclusion ordered set: :$\map f x \subseteq \map f y$ By definition of $f$: :$x^{\mathrm {compact} } \subseteq y^{\mathrm{compact} }$ By Supremum of Subset: :$\map \sup {x^{\mathrm {compact} } } \preceq \map \sup {y^{\mathrm {compact} } }$ By definition of algebraic: :$L$ satisfies axiom of K-approximation. Thus by axiom of K-approximation: :$x \preceq y$ {{qed|lemma}} We will prove that :$f$ is a surjection. Let $y \in \map {\mathit {Ids} } C$ Define $x = \sup_L y$ Thus $x \in S$. We will prove that: :$x^{\mathrm {compact} } \subseteq y$ Let $d \in x^{\mathrm {compact} }$ By definition of compact closure: :$d$ is a compact element and $d \preceq x$ By definition of compact subset: :$d \in \map K L$ By definition of compact element: :$d \ll d$ where $\ll$ denotes the way below relation. By definition of way below relation: :$\exists z \in y: d \preceq z$ By definition of ordered subset: :$d \preceq' z$ Thus by definition of lower set: :$d \in y$ {{qed|lemma}} We will prove that :$y \subseteq x^{\mathrm {compact} }$ Let $d \in y$. By definition of subset: :$d \in \map K L$ By definition of compact subset: :$d$ is a compact element. By definition of supremum: :$x$ is upper bound for $y$. By definition of upper bound: :$d \preceq x$ Thus by definition of compact closure: :$d \in x^{\mathrm {compact} }$ {{qed|lemma}} By definition of set equality: :$y = x^{\mathrm {compact} }$ Hence $y = \map f x$ {{qed|lemma}} Hence $f$ is order isomorphism. {{qed}}
We have: {{begin-eqn}} {{eqn | l = 35       | r = 5 \times 7       | c =  }} {{eqn | l = 4374       | r = 2 \times 3^7       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 36       | r = 2^2 \times 3^2       | c =  }} {{eqn | l = 4375       | r = 5^4 \times 7       | c =  }} {{end-eqn}} Thus both pairs of integers can be seen to have the same prime factors: :$2, 3, 5, 7$ {{qed}}
=== Definition 2 implies Definition 3 === Let the extremal $y = \map y x$ satisfy: :$\map y a = A$ Let the extremal $y^*_\alpha$ pass through $M = \tuple {a, A}$ and satisfy: :$\map { {y^*}'_\alpha} a - \map {y'} a = \alpha$ Then the following is a valid expression for $y^*_\alpha$: :$\map {y^*_\alpha} x = \map y x + \map h x \alpha + \epsilon$ where $\epsilon = k \alpha$ with: :$\alpha \to 0 \implies k \to 0$ and: :$\map h a = 0$ :$\map {h'} a = 1$  where $\map h x$ is an appropriate solution to Jacobi's equation. Let: :$\map h {\tilde a} = 0$ :$\beta = \sqrt {\dfrac \epsilon \alpha}$ {{finish|Explain the rest of lemma}} Then: :$\map h x \ne 0 \quad \forall x \in \openint a b \implies \map {h'} a \ne 0$ {{explain|rewrite the above unambiguously}} By Taylor's Theorem, the expression: :$\map {y_\alpha} x - \map y x = \map h x \alpha + \epsilon$ takes values with different signs at the points $\tilde a - \beta$ and $\tilde a + \beta$. Since: :$\alpha \to 0 \implies \beta \to 0$ the limit of points of intersection of $y = \map {y_\alpha^*} x$ and $y = \map y x$, as $\alpha \to 0$ is $\map {\tilde M} {\tilde a, \map y {\tilde a} }$  {{qed|lemma}} === Definition 3 implies Definition 2 === {{ProofWanted|"It is clear ..."}} {{finish|Definition 1 to link in yet}}
By the definition of an equivalence relation it is easily seen that $\set i \subseteq \eqclass i {\RR_\sigma}$. Suppose that $i \in \Fix \sigma$. Let $j \in \eqclass i {\RR_\sigma}$. Then by Condition for Membership of Equivalence Class and Permutation Induces Equivalence Relation: :$j \in \eqclass i {\RR_\sigma} \iff i \mathrel {\RR_\sigma} j \implies \exists m \in \Z: \sigma^m \paren i = j$ And by Fixed Point of Permutation is Fixed Point of Power: :$\sigma^m \paren i = i \implies i = j$ Therefore: :$\eqclass i {\RR_\sigma} \subseteq \set i$ and so: :$\eqclass i {\RR_\sigma} = \set i$ {{qed|lemma}} For the converse, suppose that: :$\eqclass i {\RR_\sigma} = \set i$. It is seen that: :$i \mathrel {\RR_\sigma} \map \sigma i \iff \map \sigma i \in \eqclass i {\RR_\sigma}$ Therefore: :$\map \sigma i = i \implies i \in \Fix \sigma$ Hence the result. {{qed}} Category:Symmetric Groups Category:Equivalence Classes pnk9h7r4rvcg60113ff3gaafm77d7y6
Let $n, k \in \Z$. Let the representations of $n$ and $k$ to the base $p$ be given by: :$n = 2^r a_r + \cdots + 2 a_1 + a_0$ :$k = 2^r b_r + \cdots + 2 b_1 + b_0$ By Corollary to Lucas' Theorem: :$\displaystyle \dbinom n k \equiv \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \pmod 2$ By definition, $a_j$ and $b_j$ are either $0$ or $1$. We have: :$\dbinom 0 1 = 0$ :$\dbinom 0 0 = \dbinom 1 0 = \dbinom 1 1 = 1$ Thus $\dbinom n k$ is odd {{iff}} none of $\dbinom {a_j} {b_j} = 0$. We are to determine how many values of $k$ there are such that $\dbinom n k$ is odd. We construct a $k$ such that $\dbinom n k$ is odd. For each $a_j = 1$, we can have $b_j = 0$ or $1$. For each $a_j = 0$, we must have $b_j = 0$. Hence for each $a_j$, there are $2^{a_j}$ ways for the corresponding $b_j$ to be such that $\dbinom {a_j} {b_j} \ne 0$. By Product Rule for Counting, the number of choices for $a_0, \dots, a_r$ for which $k$ is such that $\dbinom n k$ is odd is given by:  :$\displaystyle \prod_{j \mathop = 0}^r 2^{a_j} = 2^{\sum_{j = 0}^r {a_j} }$ which is a power of $2$. Thus the number of odd entries in a row of Pascal's triangle is a power of $2$. {{qed}}
Let $z \in x^{\mathrm{compact} }$ By definition of compact closure: :$z$ is a compact element and $z \preceq x$ By definition of transitivity: :$z \preceq y$ Thus by definition of compact closure: :$z \in y^{\mathrm{compact} }$ {{qed}}
By definition, the entries in $n$th row of Pascal's triangle are exactly the binomial coefficients: :$\dbinom n 0, \dbinom n 1, \ldots, \dbinom n n$ The result follows from Sum of Binomial Coefficients over Lower Index. {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :The sum of all the entries in the $n$th row of Pascal's triangle is equal to $2^n$. === Basis for the Induction === $\map P 0$ is the case: :The sum of all the entries in the row $0$ of Pascal's triangle is equal to $2^0 = 1$. This is true, as the only non-zero entry in row $0$ is $\dbinom 0 0$ which equals $1$. Thus $\map P 0$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :The sum of all the entries in the row $k$ of Pascal's triangle is equal to $2^k$. from which it is to be shown that: :The sum of all the entries in the row $k + 1$ of Pascal's triangle is equal to $2^{k + 1}$. === Induction Step === This is the induction step: In row $k + 1$ there are $k + 2$ entries: :$\dbinom {k + 1} 0, \dbinom {k + 1} 1, \dbinom {k + 1} 2, \ldots, \dbinom {k + 1} k, \dbinom {k + 1} {k + 1}$ From Pascal's Rule: :$\dbinom {k + 1} j = \dbinom k {j - 1} + \dbinom k j$ Thus for each of the $k + 1$ entries in row $k$: :$\dbinom k 0, \dbinom k 1, \ldots, \dbinom k k$ Element $\dbinom k j$ contributes to exactly $2$ elements of row $k + 1$ :$\dbinom {k + 1} j$ and $\dbinom {k + 1} {j + 1}$ So the sum of all the elements of row $k + 1$ is equal to $2$ times the sum of all the elements of row $k$. By the induction hypothesis, that means the sum of all the elements of row $k + 1$ is equal to $2 \times 2^k$. That is, the sum of all the entries in the row $k + 1$ of Pascal's triangle is equal to $2^{k + 1}$. So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :For all $n \in \Z_{\ge 0}$, the sum of all the entries in the $n$th row of Pascal's triangle is equal to $2^n$. {{qed}}
{{ProofWanted|Need to define the simplicial polytopic numbers and demonstrate that the $n$th simplicial polytopic number of dimension $m$ is $\dbinom m n$ or whatever the formula is.}}
From Equivalence Class of Fixed Element and Biconditional Equivalent to Biconditional of Negations: :$i \notin \Fix \sigma \iff \set i \ne \eqclass i {\RR_\sigma}$ Because the Biconditional is Transitive, it suffices to show that: :$\set i \ne \eqclass i {\RR_\sigma}$ {{iff}}: :$\eqclass i {\RR_\sigma}$ contains more than one element. Suppose that $\set i \ne \eqclass i {\RR_\sigma}$. From the definition of an equivalence relation it is seen that: :$\set i \subseteq \eqclass i {\RR_\sigma}$ And from the hypothesis: :$\set i \subset \eqclass i {\RR_\sigma}$ Therefore $\eqclass i {\RR_\sigma}$ contains more than one element. {{qed|lemma}} Conversely, if $\eqclass i {\RR_\sigma}$ contains more than one element, then it is seen that: :$\set i \ne \eqclass i {\RR_\sigma}$ {{qed}} Category:Symmetric Groups Category:Equivalence Classes p80trof9qxw8s3pzafsqody2pmfv8er
By definition, the entries in the $n$th lesser diagonal of Pascal's triangle are: :$\dbinom n 0, \dbinom {n - 1} 1, \dbinom {n - 2} 2, \dbinom {n - 3} 3, \ldots$ and so the statement can be written: :$F_{n + 1} = \displaystyle \sum_{k \mathop \ge 0} \dbinom {n - k} k$ The proof proceeds by strong induction. For all $n \in \Z_{>0}$, let $\map P n$ be the proposition: :$F_{n + 1} = \displaystyle \sum_{k \mathop \ge 0} \dbinom {n - k} k$ $\map P 0$ is the case: {{begin-eqn}} {{eqn | l = \sum_{k \mathop \ge 0} \dbinom {0 - k} k       | r = \dbinom 0 0       | c =  }} {{eqn | r = 1       | c =  }} {{eqn | r = F_1       | c =  }} {{end-eqn}} Thus $\map P 0$ is seen to hold. === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \sum_{k \mathop \ge 0} \dbinom {1 - k} k       | r = \dbinom 1 0 + \dbinom 0 1       | c =  }} {{eqn | r = 1 + 0       | c =  }} {{eqn | r = F_2       | c =  }} {{end-eqn}} Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P j$ is true, for all $j$ such that $0 \le j \le r$, then it logically follows that $\map P {r + 1}$ is true. So this is the induction hypothesis: {{begin-eqn}} {{eqn | l = F_r       | r = \sum_{k \mathop \ge 0} \dbinom {r - 1 - k} k }} {{eqn | l = F_{r + 1}       | r = \sum_{k \mathop \ge 0} \dbinom {r - k} k }} {{end-eqn}} from which it is to be shown that: :$F_{r + 2} = \displaystyle \sum_{k \mathop \ge 0} \dbinom {r + 1 - k} k$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = F_{r + 2}       | r = F_r + F_{r + 1}       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = \sum_{k \mathop \ge 0} \dbinom {r - 1 - k} k + \sum_{k \mathop \ge 0} \dbinom {r - k} k       | c = Induction Hypothesis }} {{eqn | r = \dbinom {r - 1} 0 + \dbinom {r - 2} 1 + \dbinom {r - 3} 2  + \dbinom {r - 4} 3 + \ldots       | c =  }} {{eqn | ro= +       | o =        | r = \dbinom r 0 + \dbinom {r - 1} 1 + \dbinom {r - 2} 2 + \dbinom {r - 3} 3 + \dbinom {r - 4} 4 + \ldots       | c =  }} {{eqn | r = \dbinom r 0 + \dbinom r 1 + \dbinom {r - 1} 2 + \dbinom {r - 2} 3 + \dbinom {r - 3} 4 + \ldots       | c = Pascal's Rule }} {{eqn | r = \dbinom {r + 1} 0 + \dbinom r 1 + \dbinom {r - 1} 2 + \dbinom {r - 2} 3 + \dbinom {r - 3} 4 + \ldots       | c = Binomial Coefficient with Zero }} {{eqn | r = \sum_{k \mathop \ge 0} \dbinom {r + 1 - k} k       | c =  }} {{end-eqn}} So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{>0}: F_{n + 1} = \displaystyle \sum_{k \mathop \ge 0} \dbinom {n - k} k$ {{qed}}
By definition of ordered subset: :$S_2 \subseteq S_3$ Then define $g = f:S_1 \to S_3$ Let $X$ be a subset of $S_1$ such that :$X$ admits a infimum in $L_1$. Thus by definition of complete lattice: :$g \sqbrk X$ admits a infimum in $L_3$. By definition of mapping preserves infima: :$f \sqbrk X$ admits a infimum in $L_2$ and $\map {\inf_{L_2} } {f \sqbrk X} = \map f {\map {\inf_{L_1} } X}$ Thus by definition of infima inheriting: :$\map {\inf_{L_3} } {g \sqbrk X} = \map g {\inf_{L_1} } X$ {{qed}}
By definition of ordered subset: :$S_2 \subseteq S_3$ Then define $g = f:S_1 \to S_3$ Let $X$ be a directed subset of $S_1$ such that :$X$ admits a supremum in $L_1$. Thus by definition of complete lattice: :$g\left[{X}\right]$ admits a supremum in $L_3$. By definition of mapping preserves directed suprema: :$f\left[{X}\right]$ admits a supremum in $L_2$ and $\sup_{L_2} \left({f\left[{X}\right]}\right) = f\left({\sup_{L_1} X}\right)$ By Directed Suprema Preserving Mapping is Increasing: :$f$ is an increasing mapping. By Image of Directed Subset under Increasing Mapping is Directed: :$f\left[{X}\right]$ is directed. Thus by definition of directed suprema inheriting: :$\inf_{L_3} \left({g\left[{X}\right]}\right) = g\left({\inf_{L_1} X}\right)$ {{qed}}