Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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.

For relations RR and SS and a set AA, the collections { (b,a):(a,b)∈R }\{\,(b,a) : (a,b) \in R\,\}, { (a,c):∃b ((a,b)∈R∧(b,c)∈S) }\{\,(a,c) : \exists b\ ((a,b) \in R \wedge (b,c) \in S)\,\} and { (a,b)∈R:a∈A }\{\,(a,b) \in R : a \in A\,\} are sets

Statement

Let RR and SS be relations and AA a set. Then each of

{ (b,a):(a,b)∈R },{ (a,c):∃b ((a,b)∈R∧(b,c)∈S) },{ (a,b)∈R:a∈A }\{\,(b,a) : (a,b) \in R\,\}, \qquad \{\,(a,c) : \exists b\ ((a,b) \in R \wedge (b,c) \in S)\,\}, \qquad \{\,(a,b) \in R : a \in A\,\}

is a set; the first is included in ran⁡R×dom⁡R\operatorname{ran} R \times \operatorname{dom} R, the second in dom⁡R×ran⁡S\operatorname{dom} R \times \operatorname{ran} S, and the third in RR.

Facts & Assumptions

Given: relations RR and SS and a set AA.

[L1]

A relation is a set RR every element of which is an ordered pair (Relation, dom⁡R\operatorname{dom} R, ran⁡R\operatorname{ran} R, fld⁡R\operatorname{fld} R, and the specialisations "relation from AA to BB" and "relation on AA").

[L2]

dom⁡R:={ a:∃b (a,b)∈R },ran⁡R:={ b:∃a (a,b)∈R }\operatorname{dom} R := \{\, a : \exists b\ (a,b) \in R \,\}, \qquad \operatorname{ran} R := \{\, b : \exists a\ (a,b) \in R \,\} (Relation, dom⁡R\operatorname{dom} R, ran⁡R\operatorname{ran} R, fld⁡R\operatorname{fld} R, and the specialisations "relation from AA to BB" and "relation on AA").

[L3]

z∈A×Bz \in A \times B holds if and only if z=(a,b)z = (a,b) for some a∈Aa \in A and some b∈Bb \in B (The Cartesian product A×B:={ z∈P(P(A∪B)):∃a∈A ∃b∈B z=(a,b) }A \times B := \{\, z \in \mathcal{P}(\mathcal{P}(A \cup B)) : \exists a \in A\ \exists b \in B\ z = (a,b) \,\}).

[L4]

For any parameters pˉ\bar p and any set xx, there is a set yy whose elements are exactly the elements zz of xx for which φ(z,pˉ)\varphi(z,\bar p) holds (The Axiom Schema of Separation: for each formula φ\varphi, ∀pˉ ∀x ∃y ∀z (z∈y↔(z∈x∧φ(z,pˉ)))\forall \bar p\,\forall x\,\exists y\,\forall z\,(z \in y \leftrightarrow (z \in x \wedge \varphi(z,\bar p)))).

[L5]

R⊆dom⁡R×ran⁡RR \subseteq \operatorname{dom} R \times \operatorname{ran} R (Every relation RR satisfies R⊆dom⁡R×ran⁡RR \subseteq \operatorname{dom} R \times \operatorname{ran} R, and RR is a relation from AA to BB if and only if dom⁡R⊆A\operatorname{dom} R \subseteq A and ran⁡R⊆B\operatorname{ran} R \subseteq B).

[L6]

x⊆yx \subseteq y means that every element of xx is an element of yy (Subset x⊆yx \subseteq y, proper subset x⊊yx \subsetneq y, and the separation notation { z∈x:φ(z) }\{\, z \in x : \varphi(z) \,\}).

[L7]

(a,b):={{a},{a,b}}(a,b) := \{\{a\},\{a,b\}\} (The Kuratowski ordered pair (a,b):={{a},{a,b}}(a,b) := \{\{a\},\{a,b\}\}).

Proof

technique · direct
1.1

If (a,b)∈R(a,b) \in R then a∈dom⁡Ra \in \operatorname{dom} R and b∈ran⁡Rb \in \operatorname{ran} R, so (b,a)∈ran⁡R×dom⁡R(b,a) \in \operatorname{ran} R \times \operatorname{dom} R; separating inside that product with the formula ∃a ∃b (z=(b,a)∧(a,b)∈R)\exists a\,\exists b\,(z = (b,a) \wedge (a,b) \in R) and the parameter RR therefore yields exactly the first collection.

L2L3L4L6L7
1.2

If (a,b)∈R(a,b) \in R and (b,c)∈S(b,c) \in S then a∈dom⁡Ra \in \operatorname{dom} R and c∈ran⁡Sc \in \operatorname{ran} S, so (a,c)∈dom⁡R×ran⁡S(a,c) \in \operatorname{dom} R \times \operatorname{ran} S; separating inside that product with the formula ∃a ∃c (z=(a,c)∧∃b ((a,b)∈R∧(b,c)∈S))\exists a\,\exists c\,\bigl(z = (a,c) \wedge \exists b\,((a,b) \in R \wedge (b,c) \in S)\bigr) and the parameters RR and SS yields exactly the second collection.

L2L3L4L6L7
1.3

The third collection consists of elements of RR, and separating inside RR with the formula ∃a ∃b (z=(a,b)∧a∈A)\exists a\,\exists b\,(z = (a,b) \wedge a \in A) and the parameters AA and RR yields exactly it; every element of RR is an ordered pair, so the formula selects the intended pairs.

L1L4L5L6L7
2.1

All three collections are therefore sets, with the stated inclusions.

step 1.1step 1.2step 1.3∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 19 results over 9 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