LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)

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.

If $f$ and $g$ are functions then $g \circ f$ is a function with domain $f^{- 1} [dom g]$ and $(g \circ f) (x) = g (f (x))$ there; $Δ_{A}$ is a function with $Δ_{A} (a) = a$ ; and $f \circ Δ_{A} = f = Δ_{B} \circ f$ for $f : A \to B$

Statement

Let $f$ and $g$ be functions and let $A$ , $B$ be sets. Then

(i) $g \circ f$ is a function, $dom (g \circ f) = f^{- 1} [dom g]$ , and $(g \circ f) (x) = g (f (x))$ for every $x$ in that domain;
(ii) $Δ_{A}$ is a function with $dom Δ_{A} = A$ and $Δ_{A} (a) = a$ for every $a \in A$ ;
(iii) if $f : A \to B$ then $f \circ Δ_{A} = f$ and $Δ_{B} \circ f = f$ .

Facts & Assumptions

Given: functions $f$ and $g$ , and sets $A$ , $B$ .

[L1]

A function is a relation $f$ such that $(a, b) \in f$ and $(a, c) \in f$ imply $b = c$ (A function is a relation $f$ with $(a, b) \in f$ and $(a, c) \in f$ implying $b = c$ ; $f : A \to B$ , the value $f (a)$ , domain and codomain).

[L2]

$(a, c) \in S \circ R$ holds if and only if $(a, b) \in R$ and $(b, c) \in S$ for some $b$ (The inverse relation $R^{- 1}$ , the composite $S \circ R$ , and the restriction $R ↾ A$ ).

[L3]

$dom R : = {a : \exists b (a, b) \in R}$ (Relation, $dom R$ , $ran R$ , $fld R$ , and the specialisations "relation from $A$ to $B$ " and "relation on $A$ ").

[L4]

$a \in R^{- 1} [B]$ holds if and only if $(a, b) \in R$ for some $b \in B$ (The image $R [A]$ and the preimage $R^{- 1} [B]$ of a set under a relation).

[L5]

$(a, b) \in Δ_{A}$ holds if and only if $a = b$ and $a \in A$ (The identity relation $Δ_{A} = {(a, b) \in A \times A : a = b}$ and the membership relation $\in_{A} = {(a, b) \in A \times A : a \in b}$ ).

[L6]

if $R$ is a relation from $A$ to $B$ , then $Δ_{B} \circ R = R$ and $R \circ Δ_{A} = R$ ( $T \circ (S \circ R) = (T \circ S) \circ R$ , $(S \circ R)^{- 1} = R^{- 1} \circ S^{- 1}$ , $(R^{- 1})^{- 1} = R$ , $dom (R^{- 1}) = ran R$ , and $Δ_{B} \circ R = R = R \circ Δ_{A}$ for a relation $R$ from $A$ to $B$ ).

Proof

technique · direct

1.1L1L2

Claim (i), single-valuedness: $g \circ f$ is a relation, and if $(a, c)$ and $(a, c^{'})$ both lie in it, there are $b$ and $b^{'}$ with $(a, b), (a, b^{'}) \in f$ and $(b, c), (b^{'}, c^{'}) \in g$ ; single-valuedness of $f$ gives $b = b^{'}$ , and then single-valuedness of $g$ gives $c = c^{'}$ .

1.2L1L3L5

Claim (ii): if $(a, b)$ and $(a, c)$ lie in $Δ_{A}$ then $b = a = c$ , so $Δ_{A}$ is a function; its domain is $A$ , because $(a, a) \in Δ_{A}$ exactly for $a \in A$ , and its value at $a$ is $a$ .

1.3L6

Claim (iii): a function $f : A \to B$ has $dom f = A$ and $ran f \subseteq B$ , so it is a relation from $A$ to $B$ , and the identity laws for relations apply verbatim.

2.1L1L2L3L4step 1.1

Claim (i), domain and values: $a \in dom (g \circ f)$ holds exactly when there are $b$ and $c$ with $(a, b) \in f$ and $(b, c) \in g$ , that is, exactly when $a \in dom f$ and $f (a) \in dom g$ ; and that is exactly the condition $a \in f^{- 1} [dom g]$ . For such an $a$ the pair $(a, g (f (a)))$ lies in $g \circ f$ , so $(g \circ f) (a) = g (f (a))$ by step 1.1.

If $f$ and $g$ are functions then $g \circ f$ is a function with domain $f^{- 1} [dom g]$ and $(g \circ f) (x) = g (f (x))$ there; $Δ_{A}$ is a function with $Δ_{A} (a) = a$ ; and $f \circ Δ_{A} = f = Δ_{B} \circ f$ for $f : A \to B$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

If ff and gg are functions then g∘fg∘f is a function with domain f−1[dom⁡g]f−1[domg] and (g∘f)(x)=g(f(x))(g∘f)(x)=g(f(x)) there; ΔAΔA​ is a function with ΔA(a)=aΔA​(a)=a; and f∘ΔA=f=ΔB∘ff∘ΔA​=f=ΔB​∘f for f:A→Bf:A→B

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

If $f$ and $g$ are functions then $g \circ f$ is a function with domain $f^{- 1} [dom g]$ and $(g \circ f) (x) = g (f (x))$ there; $Δ_{A}$ is a function with $Δ_{A} (a) = a$ ; and $f \circ Δ_{A} = f = Δ_{B} \circ f$ for $f : A \to B$