ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
AI-adapted — a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated — a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

In $F^{3}$ the three coordinate lines are linear subspaces whose internal direct sum is $F^{3}$ , and $F^{0}$ is the zero space

Example

Let $F$ be a field (Field) and consider the vector space $F^{3}$ of functions $3 \to F$ with the pointwise operations (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = \{0, 1, \dots, n-1\}$ ). Since $3 = \{0, 1, 2\}$ (The natural numbers $\mathbb{N}$ (von Neumann), On $\mathbb{N}$ the order is membership: $m < n \iff m \in n$ ), an element is written $x = (x_0, x_1, x_2)$ with $x_i := x(i)$ , indexed from $0$ . For $j < 3$ let $e_j \in F^{3}$ be given by $e_j(j) = 1_F$ and $e_j(i) = 0_F$ for $i \ne j$ , and put

$L_j \;:=\; \{\, x \in F^{3} \;:\; x_i = 0_F \text{ for every } i < 3 \text{ with } i \ne j \,\}.$

Then:

$L_j = \{\, \lambda e_j : \lambda \in F \,\} = \operatorname{span}\{e_j\}$ , so each $L_j$ is a linear subspace of $F^{3}$ (Linear subspace of a vector space, Linear combination of a finite list, and the span $\operatorname{span}(S)$ as the smallest linear subspace containing $S$ );
$F^{3} = \bigoplus_{j<3} L_j$ (Internal direct sum $V = \bigoplus_{i<n} U_i$ : the sum is everything and each summand meets the sum of the others only in $0_V$ ), the unique decomposition of $x \in F^{3}$ being $x = x_0 e_0 + x_1 e_1 + x_2 e_2$ ;
$F^{0}$ is the zero space: it has exactly one element, the empty function.

The sets $L_j$ are called the coordinate lines of $F^{3}$ ; the word "line" is used informally, since dimension is not available here and nothing below uses it.

Facts & Assumptions

Given: A field $F$ , the vector space $F^{3}$ of functions $3 \to F$ with pointwise operations, the vectors $e_j$ for $j < 3$ , and the sets $L_j$ as displayed.

[L1]

$F^{X}$ is a vector space over $F$ with $(x+y)(i) = x(i)+y(i)$ , $(\lambda x)(i) = \lambda x(i)$ and zero the constant function at $0_F$ ; for $X = n$ a natural number, $n = \{0,\dots,n-1\}$ ; and $F^{0}$ has exactly one element, the empty function, which is its zero vector (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = \{0, 1, \dots, n-1\}$ , Vector space over a field, The natural numbers $\mathbb{N}$ (von Neumann), On $\mathbb{N}$ the order is membership: $m < n \iff m \in n$ ).

[L2]

$\operatorname{span}\{v\} = \{\, \lambda v : \lambda \in F \,\}$ , and the span of a subset is a linear subspace ( $\operatorname{span}\{v\} = \{\, \lambda v : \lambda \in F \,\}$ , which is $\{0_V\}$ when $v = 0_V$ , and when $v \ne 0_V$ contains $0_V$ only as the multiple $0_F v$ , Linear combination of a finite list, and the span $\operatorname{span}(S)$ as the smallest linear subspace containing $S$ , Linear subspace of a vector space).

[L3]

The elements of $\sum_{i<n} U_i$ are exactly the $\sum_{i<n} u_i$ with $u_i \in U_i$ ; and $\sum_{i<3} u_i = (u_0 + u_1) + u_2$ , by the recursion $\sum_{i<0} u_i = 0$ and $\sum_{i<\sigma(m)} u_i = \bigl(\sum_{i<m} u_i\bigr) + u_m$ together with $0 + u_0 = u_0$ (The sum $U + W$ of two linear subspaces and the sum $\sum_{i<n} U_i$ of a finite family).

[L4]

$V = \bigoplus_{i<n} U_i$ holds if and only if every $v \in V$ is $\sum_{i<n} u_i$ with $u_i \in U_i$ in exactly one way ( $V = \bigoplus_{i<n} U_i$ if and only if every $v \in V$ is $\sum_{i<n} u_i$ with $u_i \in U_i$ in exactly one way; equivalently, if and only if the sum is $V$ and $\sum_{i<n} u_i = 0_V$ with $u_i \in U_i$ forces every $u_i = 0_V$ , Internal direct sum $V = \bigoplus_{i<n} U_i$ : the sum is everything and each summand meets the sum of the others only in $0_V$ ).

[L5]

In a field: $1_F \lambda = \lambda$ , multiplication is commutative, $0_F \lambda = 0_F$ (Multiplication by zero: $0 \cdot a = 0$ ) and hence $\lambda 0_F = 0_F$ , and $0_F$ is the additive identity (Field).

Verification

technique · direct

1.1

$F^{3}$ is the set of functions $3 \to F$ with the pointwise operations, and $3 = \{0,1,2\}$ , so the coordinates of an element are $x_0, x_1, x_2$ .

1.2

Each $e_j$ is an element of $F^{3}$ , and each $L_j$ is a subset of $F^{3}$ , both by their displayed descriptions.

L1L5

1.3

$L_j = \{\, \lambda e_j : \lambda \in F \,\}$ for every $j < 3$ . If $\lambda \in F$ then $(\lambda e_j)_j = \lambda 1_F = \lambda$ and $(\lambda e_j)_i = \lambda 0_F = 0_F$ for $i \ne j$ , so $\lambda e_j \in L_j$ . Conversely if $x \in L_j$ then $x$ and $x_j e_j$ have the same value at every $i < 3$ , namely $x_j$ at $i = j$ and $0_F$ elsewhere, so $x = x_j e_j$ .

L1L5

1.4

For $u_0, u_1, u_2 \in F^{3}$ the finite sum is $\sum_{j<3} u_j = (u_0 + u_1) + u_2$ , whose value at $i < 3$ is $(u_0(i) + u_1(i)) + u_2(i)$ , by the pointwise definition of the addition.

L1L3

1.5

$F^{0}$ has exactly one element, the empty function, and that element is its zero vector, so $F^{0}$ is the zero space; this is claim 3.

2.1

Each $L_j$ is a linear subspace of $F^{3}$ and equals $\operatorname{span}\{e_j\}$ : by step 1.3 it is the set of scalar multiples of $e_j$ , which is exactly the span of $\{e_j\}$ , and a span is a linear subspace. This is claim 1.

step 1.3L2

2.2

Every $x \in F^{3}$ decomposes. Put $u_j := x_j e_j$ , which lies in $L_j$ by step 1.3. The value of $\sum_{j<3} u_j$ at $i < 3$ is $(x_0 e_0(i) + x_1 e_1(i)) + x_2 e_2(i)$ ; since $e_j(i) = 0_F$ for $j \ne i$ and $e_i(i) = 1_F$ , exactly one summand is $x_i$ and the others are $0_F$ , so the value is $x_i$ . Hence $\sum_{j<3} u_j = x$ , and $\sum_{j<3} L_j = F^{3}$ .

step 1.3step 1.4L1L5

3.1

The decomposition is unique. Suppose $u_j \in L_j$ for $j < 3$ and $\sum_{j<3} u_j = x$ . Evaluating at $i < 3$ gives $(u_0(i) + u_1(i)) + u_2(i) = x_i$ , and $u_j(i) = 0_F$ whenever $j \ne i$ , so the left-hand side is $u_i(i)$ ; thus $u_i(i) = x_i$ , and step 1.3 gives $u_i = u_i(i) e_i = x_i e_i$ . So the list is the one of step 2.2.

step 1.3step 1.4L1L5

4.1

By steps 2.2 and 3.1 every $x \in F^{3}$ is $\sum_{j<3} u_j$ with $u_j \in L_j$ in exactly one way, so $F^{3} = \bigoplus_{j<3} L_j$ , and the decomposition is $x = x_0 e_0 + x_1 e_1 + x_2 e_2$ . This is claim 2.

step 2.2step 3.1L4

5.1

Claim 1 is step 2.1, claim 2 is step 4.1 and claim 3 is step 1.5.

step 1.5step 2.1step 4.1∎

Remarks

Every index here starts at $0$ . The coordinates are $x_0, x_1, x_2$ and the summands are $L_0, L_1, L_2$ , because $3 = \{0,1,2\}$ as a von Neumann natural. Writing the same example with coordinates $x_1, x_2, x_3$ would not match the definition of $F^{n}$ used here (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = \{0, 1, \dots, n-1\}$ ).
$F^{0}$ is not empty. There is exactly one function $\varnothing \to F$ , so $F^{0}$ has exactly one element and is the zero space. It is also the internal direct sum of the empty family of its linear subspaces (Internal direct sum $V = \bigoplus_{i<n} U_i$ : the sum is everything and each summand meets the sum of the others only in $0_V$ ), which is the only space that is.
Nothing above is special to $3$ . The same computation works for any $n \in \mathbb{N}$ and gives $F^{n} = \bigoplus_{j<n} L_j$ , at $n = 0$ degenerating to the statement that the zero space is the direct sum of the empty family. It is written out at $n = 3$ so that the finite sums are the explicit $(u_0 + u_1) + u_2$ rather than an induction.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 59 results over 22 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Examples of vector spaces (Wikipedia) (standard reference, not scraped)
Direct sum of modules (Wikipedia) (standard reference, not scraped)
S. Axler, Linear Algebra Done Right, 4th ed. (free PDF, CC BY-NC) (standard reference, not scraped)

In F3F^{3}F3 the three coordinate lines are linear subspaces whose internal direct sum is F3F^{3}F3, and F0F^{0}F0 is the zero space