Alphabeta Math
Session-authored (Fable 5 assisted)
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.

19 results · all verified · 0 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full by a delegated reviewing agent on the owner's instruction; the judge is an additional, independent cross-model AI review of the proofs. The 19 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

The ZFC Axioms and the Basic Set Constructions

1 · Prerequisites

None. This page is self-contained.

2 · Summary

This development starts from first-order logic with equality over a single binary relation symbol \in, in which every object of the domain is a set and the domain is nonempty; nothing mathematical is assumed before that. The one result taken from elsewhere is recorded rather than proved: neither the continuum hypothesis nor its generalisation is decided by these axioms (The continuum hypothesis and its generalisation are independent of ZFC ), and it is cited only to say what the Power Set axiom leaves open about the size of a power set.

The ZF axioms are stated in that language, written without abbreviations wherever the usual formulation would need notation introduced further down. The Axiom of Choice is the one axiom of ZFC not stated here; this library states it at The Axiom of Choice. Separation and Extensionality yield the empty set and its uniqueness; Pairing, Union, Separation and Power Set then give unordered pairs, unions, intersections, differences and power sets, with the commutative, associative, distributive and De Morgan laws proved from the resulting membership criteria. Russell's paradox rules out a set of all sets and with it an intersection of the empty family, Foundation rules out short membership cycles and makes xx{x}x \mapsto x \cup \{x\} injective, and the Kuratowski ordered pair and its characterising property produce the Cartesian product.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-08-06 (claude-opus-5)Open item page →

The first-order language of set theory: \in, ==, formulas with parameters, and class abbreviations

Definition

The language of set theory is first-order logic with equality whose only non-logical symbol is a binary relation symbol \in. Its terms are the variables x,y,z,x, y, z, \dots; its atomic formulas are xyx \in y and x=yx = y; and its formulas are built from atomic formulas by the connectives ¬\neg, \wedge, \vee, \to, \leftrightarrow and the quantifiers x\forall x and x\exists x. Each axiom stated on this page is a sentence of this language, and no other vocabulary occurs in any of them.

Two conventions of ordinary first-order logic are in force throughout.

  • The domain of discourse is nonempty: x(x=x)\exists x\,(x = x) is logically valid, so some object exists before any axiom is assumed.
  • Every object of the domain is a set. There are no urelements, so "zz" and "the set zz" say the same thing.

We write xyx \neq y for ¬(x=y)\neg(x = y) and xyx \notin y for ¬(xy)\neg(x \in y); we abbreviate z(zxφ)\forall z\,(z \in x \to \varphi) by zxφ\forall z \in x\,\varphi and z(zxφ)\exists z\,(z \in x \wedge \varphi) by zxφ\exists z \in x\,\varphi; and we abbreviate w(ψ(w)v(ψ(v)v=w))\exists w\,\bigl(\psi(w) \wedge \forall v\,(\psi(v) \to v = w)\bigr) by !wψ(w)\exists! w\,\psi(w), read "there is exactly one ww with ψ(w)\psi(w)".

A formula is written φ(z,pˉ)\varphi(z, \bar p) when its free variables are among zz and the variables of the list pˉ=p1,,pn\bar p = p_1, \dots, p_n; the pip_i are its parameters. Parameters matter because an axiom schema ranges over formulas, and a formula may mention sets already in hand.

A class abbreviation {z:φ(z,pˉ)}\{\, z : \varphi(z, \bar p) \,\} is not a term of the language. It is notation for the formula φ\varphi itself, and a sentence written with it abbreviates a sentence that does not use it:

  • w{z:φ(z,pˉ)}w \in \{\, z : \varphi(z, \bar p) \,\} abbreviates φ(w,pˉ)\varphi(w, \bar p);
  • "{z:φ(z,pˉ)}\{\, z : \varphi(z, \bar p) \,\} is a set" abbreviates yz(zyφ(z,pˉ))\exists y\,\forall z\,\bigl(z \in y \leftrightarrow \varphi(z, \bar p)\bigr).

Every formula has a class abbreviation, and whether that abbreviation is a set is a separate question; answering it affirmatively always requires an axiom.

Remarks

DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-08-06 (claude-opus-5)Open item page →

The Axiom of Extensionality: xy(z(zxzy)x=y)\forall x\,\forall y\,(\forall z\,(z \in x \leftrightarrow z \in y) \to x = y)

Definition

The Axiom of Extensionality is the sentence

xy(z(zxzy)x=y)\forall x\,\forall y\,\bigl(\forall z\,(z \in x \leftrightarrow z \in y) \to x = y\bigr)

of the language of set theory (The first-order language of set theory: \in, ==, formulas with parameters, and class abbreviations): if every zz satisfies zxz \in x if and only if zyz \in y, then x=yx = y.

A set is therefore determined by its members and by nothing else. Order, repetition and any description used to present a set are invisible to identity.

Remarks

  • The converse is logic, not an axiom. If x=yx = y then zxz \in x and zyz \in y are the same statement for every zz, by the substitution rule for equality. So Extensionality upgrades to the biconditional x=yz(zxzy)x = y \leftrightarrow \forall z\,(z \in x \leftrightarrow z \in y), and only the right-to-left direction of that biconditional — the implication from sameness of members to equality — is assumed.

  • What it is used for below. Every construction on this page produces a set by an axiom that specifies its members, and every such axiom is stated with \leftrightarrow. Extensionality is what turns "a set with these members" into "the set with these members", so it is the licence for each definite article and each piece of notation introduced here.

DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-08-06 (claude-opus-5)Open item page →

The Axiom Schema of Separation: for each formula φ\varphi, pˉxyz(zy(zxφ(z,pˉ)))\forall \bar p\,\forall x\,\exists y\,\forall z\,(z \in y \leftrightarrow (z \in x \wedge \varphi(z,\bar p)))

Definition

Let φ(z,pˉ)\varphi(z, \bar p) be a formula of the language of set theory (The first-order language of set theory: \in, ==, formulas with parameters, and class abbreviations) in which the variable yy does not occur free. The Separation instance for φ\varphi is the sentence

pˉxyz(zy(zxφ(z,pˉ))).\forall \bar p\,\forall x\,\exists y\,\forall z\,\bigl(z \in y \leftrightarrow (z \in x \wedge \varphi(z,\bar p))\bigr).

The Axiom Schema of Separation is the collection of all these sentences, one for each such φ\varphi. In words: 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.

It is a schema and not a single axiom because φ\varphi ranges over the formulas of the language, of which there are infinitely many, and the language provides no way to quantify over them.

Remarks

DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-08-06 (claude-opus-5)Open item page →

The Axiom of Pairing: xyzt(tz(t=xt=y))\forall x\,\forall y\,\exists z\,\forall t\,(t \in z \leftrightarrow (t = x \vee t = y))

Definition

The Axiom of Pairing is the sentence

xyzt(tz(t=xt=y))\forall x\,\forall y\,\exists z\,\forall t\,\bigl(t \in z \leftrightarrow (t = x \vee t = y)\bigr)

of the language of set theory (The first-order language of set theory: \in, ==, formulas with parameters, and class abbreviations): for any xx and yy there is a set zz whose elements are exactly xx and yy.

Remarks

DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-08-06 (claude-opus-5)Open item page →

The Axiom of Union: xyz(zys(sxzs))\forall x\,\exists y\,\forall z\,(z \in y \leftrightarrow \exists s\,(s \in x \wedge z \in s))

Definition

The Axiom of Union is the sentence

xyz(zys(sxzs))\forall x\,\exists y\,\forall z\,\bigl(z \in y \leftrightarrow \exists s\,(s \in x \wedge z \in s)\bigr)

of the language of set theory (The first-order language of set theory: \in, ==, formulas with parameters, and class abbreviations): for any set xx there is a set yy whose elements are exactly the sets zz belonging to some member ss of xx.

The axiom removes one layer of membership: it collects the members of the members of xx, not the members of xx.

Remarks

DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-08-06 (claude-opus-5)Open item page →

The Axiom of Power Set: xyz(t(tztx)zy)\forall x\,\exists y\,\forall z\,(\forall t\,(t \in z \to t \in x) \to z \in y)

Definition

The Axiom of Power Set is the sentence

xyz(t(tztx)zy)\forall x\,\exists y\,\forall z\,\bigl(\forall t\,(t \in z \to t \in x) \to z \in y\bigr)

of the language of set theory (The first-order language of set theory: \in, ==, formulas with parameters, and class abbreviations): for every set xx there is a set yy that contains every zz all of whose elements belong to xx.

The axiom is assumed here in this implication form. It says that some set collects all such zz; it does not say that yy contains nothing else, and trimming yy down to exactly those zz is a separate step, carried out at For every set xx there is exactly one set whose elements are precisely the subsets of xx using The Axiom Schema of Separation: for each formula φ\varphi, pˉxyz(zy(zxφ(z,pˉ)))\forall \bar p\,\forall x\,\exists y\,\forall z\,(z \in y \leftrightarrow (z \in x \wedge \varphi(z,\bar p))).

Remarks

DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-08-06 (claude-opus-5)Open item page →

The Axiom Schema of Replacement: for each formula φ\varphi, if φ\varphi defines a class function on AA then its image on AA is a set

Definition

Let φ(z,w,pˉ)\varphi(z, w, \bar p) be a formula of the language of set theory (The first-order language of set theory: \in, ==, formulas with parameters, and class abbreviations) in which AA and BB do not occur free. The Replacement instance for φ\varphi is the sentence

pˉA(z(zA!wφ(z,w,pˉ))Bw(wBz(zAφ(z,w,pˉ)))).\forall \bar p\,\forall A\,\Bigl(\forall z\,\bigl(z \in A \to \exists! w\,\varphi(z,w,\bar p)\bigr) \to \exists B\,\forall w\,\bigl(w \in B \leftrightarrow \exists z\,(z \in A \wedge \varphi(z,w,\bar p))\bigr)\Bigr).

The Axiom Schema of Replacement is the collection of all these sentences, one for each such φ\varphi. In words: if for every zAz \in A there is exactly one ww with φ(z,w,pˉ)\varphi(z,w,\bar p), then there is a set BB whose elements are exactly those ww.

Remarks

DefinitionDefinition: AI-adaptedProof: Not applicableverified 2026-08-06 (claude-opus-5)Open item page →

The Axiom of Infinity: there is a set containing a set with no elements and closed under yy{y}y \mapsto y \cup \{y\}

Definition

The Axiom of Infinity is the sentence

I(e(eI¬t(te))y(yIs(sIt(ts(tyt=y)))))\exists I\,\Bigl(\exists e\,\bigl(e \in I \wedge \neg\exists t\,(t \in e)\bigr) \wedge \forall y\,\bigl(y \in I \to \exists s\,(s \in I \wedge \forall t\,(t \in s \leftrightarrow (t \in y \vee t = y)))\bigr)\Bigr)

of the language of set theory (The first-order language of set theory: \in, ==, formulas with parameters, and class abbreviations).

It is written here in \in and == alone, with no abbreviations, because the notation it is usually stated in is introduced later on this page. Once that notation is available, the sentence reads: there is a set II with I\varnothing \in I such that yIy \in I implies y{y}Iy \cup \{y\} \in I. The first conjunct is I\varnothing \in I written out, and the inner clause t(ts(tyt=y))\forall t\,(t \in s \leftrightarrow (t \in y \vee t = y)) says exactly that ss is y{y}y \cup \{y\}.

Remarks

DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-08-06 (claude-opus-5)Open item page →

The Axiom of Foundation: S(t(tS)s(sS¬u(usuS)))\forall S\,(\exists t\,(t \in S) \to \exists s\,(s \in S \wedge \neg\exists u\,(u \in s \wedge u \in S)))

Definition

The Axiom of Foundation, also called the Axiom of Regularity, is the sentence

S(t(tS)s(sS¬u(usuS)))\forall S\,\Bigl(\exists t\,(t \in S) \to \exists s\,\bigl(s \in S \wedge \neg\exists u\,(u \in s \wedge u \in S)\bigr)\Bigr)

of the language of set theory (The first-order language of set theory: \in, ==, formulas with parameters, and class abbreviations): every set with at least one member has a member ss that shares no member with it.

It is written here without abbreviations, for the same reason as The Axiom of Infinity: there is a set containing a set with no elements and closed under yy{y}y \mapsto y \cup \{y\}: the usual statement uses notation introduced later on this page. In that notation it reads: every nonempty set SS has a member ss with sS=s \cap S = \varnothing. Such an ss is called an \in-minimal member of SS.

Remarks

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

There is exactly one set with no elements, written \varnothing

Statement

There is exactly one set with no elements: there is a set ee such that ¬z(ze)\neg\exists z\,(z \in e), and any two such sets are equal. That set is written \varnothing.

Facts & Assumptions

Proof

technique · direct
1.1

The domain of discourse is nonempty, so fix a set aa.

givenchoose
2.1

Apply Separation to aa with the formula φ(z):=¬(z=z)\varphi(z) := \neg(z = z): there is a set ee such that, for every zz, zez \in e holds if and only if zaz \in a and ¬(z=z)\neg(z = z).

L1step 1.1
3.1

No zz satisfies ¬(z=z)\neg(z = z), so no zz satisfies zez \in e; hence ee is a set with no elements, which proves existence.

step 2.1
4.1

If ee' is also a set with no elements, then zez \in e and zez \in e' both fail for every zz, so zez \in e holds if and only if zez \in e', and therefore e=ee = e'; existence and uniqueness together give the statement, and \varnothing denotes this set.

L2step 3.1

Remarks

  • Existence is derived, not assumed. Several presentations take "there is a set with no elements" as an axiom of its own. Here it is a theorem, because the nonemptiness of the domain of discourse is already a validity of first-order logic and Separation converts any set whatever into this one.
DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-08-06 (claude-opus-5)Open item page →

Subset xyx \subseteq y, proper subset xyx \subsetneq y, and the separation notation {zx:φ(z)}\{\, z \in x : \varphi(z) \,\}

Definition

For sets xx and yy we write xyx \subseteq y, and say that xx is a subset of yy or that xx is included in yy, for the formula t(txty)\forall t\,(t \in x \to t \in y) of the language of set theory (The first-order language of set theory: \in, ==, formulas with parameters, and class abbreviations); thus xyx \subseteq y means that every element of xx is an element of yy. We also write yxy \supseteq x for xyx \subseteq y.

xx is a proper subset of yy, written xyx \subsetneq y, when xyx \subseteq y and xyx \neq y.

Separation notation. Let xx be a set, φ(z,pˉ)\varphi(z,\bar p) a formula and pˉ\bar p parameters. The Axiom Schema of Separation: for each formula φ\varphi, pˉxyz(zy(zxφ(z,pˉ)))\forall \bar p\,\forall x\,\exists y\,\forall z\,(z \in y \leftrightarrow (z \in x \wedge \varphi(z,\bar p))) supplies a set whose elements are exactly the elements zz of xx for which φ(z,pˉ)\varphi(z,\bar p) holds, and The Axiom of Extensionality: xy(z(zxzy)x=y)\forall x\,\forall y\,(\forall z\,(z \in x \leftrightarrow z \in y) \to x = y) shows there is only one such set. It is written

{zx:φ(z,pˉ)}\{\, z \in x : \varphi(z,\bar p) \,\}

and every set introduced on this page by separating a condition inside a set already in hand is written this way. Directly from the definition, {zx:φ(z,pˉ)}x\{\, z \in x : \varphi(z,\bar p) \,\} \subseteq x.

Remarks

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

x\varnothing \subseteq x, xxx \subseteq x, inclusion is transitive, and x=yx = y if and only if xyx \subseteq y and yxy \subseteq x

Statement

For all sets xx, yy and zz:

  • (i) x\varnothing \subseteq x;
  • (ii) xxx \subseteq x;
  • (iii) if xyx \subseteq y and yzy \subseteq z then xzx \subseteq z;
  • (iv) x=yx = y if and only if xyx \subseteq y and yxy \subseteq x.

Facts & Assumptions

Given: sets xx, yy and zz.

[L3]

There is exactly one set with no elements, written \varnothing (There is exactly one set with no elements, written \varnothing).

Proof

technique · direct
1.1

Claim (i): no tt satisfies tt \in \varnothing, so the implication "tt \in \varnothing implies txt \in x" holds vacuously for every tt, which is x\varnothing \subseteq x.

L1L3
1.2

Claim (ii): every tt with txt \in x satisfies txt \in x, which is xxx \subseteq x.

L1
1.3

Claim (iii): assume xyx \subseteq y and yzy \subseteq z, and let txt \in x; then tyt \in y by the first inclusion and tzt \in z by the second, so every element of xx is an element of zz.

L1
1.4

Claim (iv), from right to left: assume xyx \subseteq y and yxy \subseteq x; for any tt, the first inclusion gives that txt \in x implies tyt \in y and the second gives that tyt \in y implies txt \in x, so txt \in x holds if and only if tyt \in y, and therefore x=yx = y.

L1L2
1.5

Claim (iv), from left to right: assume x=yx = y; then txt \in x and tyt \in y are the same statement for every tt, so each of xyx \subseteq y and yxy \subseteq x holds.

L1
2.1

Claims (i) to (iv) are established, which is the statement.

step 1.1step 1.2step 1.3step 1.4step 1.5
DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-08-06 (claude-opus-5)Open item page →

The unordered pair {x,y}\{x,y\} and the singleton {x}={x,x}\{x\} = \{x,x\}

Definition

Let xx and yy be sets. The Axiom of Pairing: xyzt(tz(t=xt=y))\forall x\,\forall y\,\exists z\,\forall t\,(t \in z \leftrightarrow (t = x \vee t = y)) gives a set whose elements are exactly xx and yy, and The Axiom of Extensionality: xy(z(zxzy)x=y)\forall x\,\forall y\,(\forall z\,(z \in x \leftrightarrow z \in y) \to x = y) shows there is only one such set; it is written {x,y}\{x,y\}. Thus {x,y}\{x,y\} is the set whose elements are exactly xx and yy, and {x}:={x,x}\{x\} := \{x,x\}, the singleton of xx, is the set whose only element is xx:

t{x,y}(t=xt=y),t{x}t=x.t \in \{x,y\} \leftrightarrow (t = x \vee t = y), \qquad t \in \{x\} \leftrightarrow t = x .

Remarks

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

{a,b}={c,d}\{a,b\} = \{c,d\} if and only if (a=ca = c and b=db = d) or (a=da = d and b=cb = c); in particular {a}={c,d}\{a\} = \{c,d\} if and only if a=c=da = c = d

Statement

For all sets aa, bb, cc, dd:

{a,b}={c,d}  ((a=cb=d)(a=db=c)).\{a,b\} = \{c,d\} \ \leftrightarrow\ \bigl((a = c \wedge b = d) \vee (a = d \wedge b = c)\bigr).

In particular, taking b:=ab := a, {a}={c,d}\{a\} = \{c,d\} if and only if a=ca = c and a=da = d.

Facts & Assumptions

Given: sets aa, bb, cc, dd.

[L1]

{x,y}\{x,y\} is the set whose elements are exactly xx and yy, and {x}:={x,x}\{x\} := \{x,x\} (The unordered pair {x,y}\{x,y\} and the singleton {x}={x,x}\{x\} = \{x,x\}).

Proof

technique · cases
1.1

Right to left: if a=ca = c and b=db = d, or a=da = d and b=cb = c, then for every tt the conditions "t=at = a or t=bt = b" and "t=ct = c or t=dt = d" are the same, so the two pairs have the same elements and are equal.

L1L2
1.2

Left to right, setting up: assume {a,b}={c,d}\{a,b\} = \{c,d\}. Since aa and bb are elements of the left-hand set they are elements of the right-hand one, so aa is cc or dd and bb is cc or dd; likewise cc is aa or bb and dd is aa or bb.

L1
2.1

Suppose a=ba = b. Then cc is aa or bb, so c=ac = a; and dd is aa or bb, so d=ad = a. Hence a=ca = c and b=a=db = a = d, the first disjunct.

assume-case samestep 1.2
2.2

Suppose instead aba \neq b. If a=ca = c, then bb is cc or dd, and b=c=ab = c = a is excluded, so b=db = d and the first disjunct holds. Otherwise a=da = d; then cc is aa or bb, and c=ac = a would give c=a=dc = a = d, whence bb, being cc or dd, equals aa, contradicting aba \neq b; so c=bc = b and the second disjunct holds.

assume-case diffstep 1.2
3.1

The alternatives a=ba = b and aba \neq b exhaust the possibilities, so the left-to-right implication holds.

step 2.1step 2.2cases-exhaustive
4.1

Both implications hold, which is the biconditional; putting b:=ab := a makes the two disjuncts read "a=ca = c and a=da = d" and "a=da = d and a=ca = c", so {a}={c,d}\{a\} = \{c,d\} if and only if a=ca = c and a=da = d.

step 1.1step 3.1L1
DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-08-06 (claude-opus-5)Open item page →

The union x\bigcup x of a set, and the binary union ab:={a,b}a \cup b := \bigcup \{a,b\}

Definition

Let xx be a set. The Axiom of Union: xyz(zys(sxzs))\forall x\,\exists y\,\forall z\,(z \in y \leftrightarrow \exists s\,(s \in x \wedge z \in s)) gives a set whose elements are exactly the sets zz belonging to some member of xx, and The Axiom of Extensionality: xy(z(zxzy)x=y)\forall x\,\forall y\,(\forall z\,(z \in x \leftrightarrow z \in y) \to x = y) shows there is only one such set; it is written x\bigcup x. Thus x\bigcup x is the set whose elements are exactly the elements of the elements of xx, and ab:={a,b}a \cup b := \bigcup\{a,b\} is the binary union of aa and bb, formed from the unordered pair of The unordered pair {x,y}\{x,y\} and the singleton {x}={x,x}\{x\} = \{x,x\}:

zxs(sxzs).z \in \bigcup x \leftrightarrow \exists s\,(s \in x \wedge z \in s).

The membership criterion for the binary union, that zabz \in a \cup b holds exactly when zaz \in a or zbz \in b, is proved at =\bigcup \varnothing = \varnothing, {a}=a\bigcup \{a\} = a, {a,b}=ab\bigcup \{a,b\} = a \cup b, {a}=a\bigcap \{a\} = a, and {a,b}=ab\bigcap \{a,b\} = a \cap b rather than assumed here.

Remarks

  • One layer only. x\bigcup x collects the elements of the elements of xx, not the elements of xx. For a set of sets of sets the operation therefore has to be applied twice, and that iteration is what makes R\bigcup\bigcup R the right ambient set for the domain and range of a relation.

  • Notation. \bigcup with a single set argument is the primitive operation supplied by the axiom; \cup between two sets is the derived binary one. The indexed form iIAi\bigcup_{i \in I} A_i is a further abbreviation, defined once indexed families are available.

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

For a set xx \neq \varnothing the collection {z:s(sxzs)}\{\, z : \forall s\,(s \in x \to z \in s) \,\} is a set, and it does not depend on the member of xx used to separate it

Statement

Let xx be a set with xx \neq \varnothing. Then there is a set whose elements are exactly the sets belonging to every member of xx; that is, the class {z:s(sxzs)}\{\, z : \forall s\,(s \in x \to z \in s) \,\} is a set. Moreover, for every BxB \in x the separated set {zB:s(sxzs)}\{\, z \in B : \forall s\,(s \in x \to z \in s) \,\} is that same set, so the construction does not depend on which member of xx is used.

Facts & Assumptions

Given: a set xx with xx \neq \varnothing.

[L3]

There is exactly one set with no elements, written \varnothing (There is exactly one set with no elements, written \varnothing).

Proof

technique · direct
1.1

If xx had no members it would be a set with no elements and hence equal to \varnothing, contrary to hypothesis; so xx has a member, and we fix one, BxB \in x.

L3givenchoose
2.1

Apply Separation to BB with the formula φ(z,x):=s(sxzs)\varphi(z,x) := \forall s\,(s \in x \to z \in s) and the parameter xx: the collection cB:={zB:s(sxzs)}c_B := \{\, z \in B : \forall s\,(s \in x \to z \in s) \,\} is a set, and cBBc_B \subseteq B.

L1L4step 1.1
3.1

For every zz, zcBz \in c_B holds exactly when zBz \in B and zz belongs to every member of xx; since BB is itself a member of xx, the second condition already forces zBz \in B, so zcBz \in c_B holds exactly when zz belongs to every member of xx.

step 2.1step 1.1
4.1

The condition characterising the elements of cBc_B in step 3.1 does not mention BB, so for any other member BB' of xx the set cBc_{B'} has exactly the same elements as cBc_B and equals it; the class is therefore a set and is independent of the member used to separate it.

L2step 3.1
DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-08-06 (claude-opus-5)Open item page →

The intersection x\bigcap x of a nonempty set, the binary intersection ab:={a,b}a \cap b := \bigcap\{a,b\}, and disjointness

Definition

Let xx be a set with xx \neq \varnothing. For a set xx \neq \varnothing the collection {z:s(sxzs)}\{\, z : \forall s\,(s \in x \to z \in s) \,\} is a set, and it does not depend on the member of xx used to separate it shows that the sets belonging to every member of xx form a set, and that it does not depend on the member of xx used to build it; it is written x\bigcap x. Thus for xx \neq \varnothing, x\bigcap x is the set whose elements are exactly the sets belonging to every element of xx, and ab:={a,b}a \cap b := \bigcap\{a,b\} is the binary intersection of aa and bb:

zxs(sxzs)(x).z \in \bigcap x \leftrightarrow \forall s\,(s \in x \to z \in s) \quad (x \neq \varnothing).

The binary case is legitimate because a{a,b}a \in \{a,b\}, so the unordered pair of The unordered pair {x,y}\{x,y\} and the singleton {x}={x,x}\{x\} = \{x,x\} is never empty. Two sets aa and bb are disjoint when ab=a \cap b = \varnothing (There is exactly one set with no elements, written \varnothing).

Remarks

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

=\bigcup \varnothing = \varnothing, {a}=a\bigcup \{a\} = a, {a,b}=ab\bigcup \{a,b\} = a \cup b, {a}=a\bigcap \{a\} = a, and {a,b}=ab\bigcap \{a,b\} = a \cap b

Statement

For all sets aa and bb:

  • (i) =\bigcup \varnothing = \varnothing;
  • (ii) {a}=a\bigcup \{a\} = a;
  • (iii) {a,b}=ab\bigcup \{a,b\} = a \cup b, and zabz \in a \cup b holds if and only if zaz \in a or zbz \in b;
  • (iv) {a}=a\bigcap \{a\} = a;
  • (v) {a,b}=ab\bigcap \{a,b\} = a \cap b, and zabz \in a \cap b holds if and only if zaz \in a and zbz \in b.

The equalities in (iii) and (v) are the definitions of aba \cup b and aba \cap b written out; what is proved about them here is the membership criterion beside each.

Facts & Assumptions

Given: sets aa and bb.

[L1]

x\bigcup x is the set whose elements are exactly the elements of the elements of xx, and ab:={a,b}a \cup b := \bigcup\{a,b\} (The union x\bigcup x of a set, and the binary union ab:={a,b}a \cup b := \bigcup \{a,b\}).

[L2]

For xx \neq \varnothing, x\bigcap x is the set whose elements are exactly the sets belonging to every element of xx, and ab:={a,b}a \cap b := \bigcap\{a,b\} (The intersection x\bigcap x of a nonempty set, the binary intersection ab:={a,b}a \cap b := \bigcap\{a,b\}, and disjointness).

[L3]

{x,y}\{x,y\} is the set whose elements are exactly xx and yy, and {x}:={x,x}\{x\} := \{x,x\} (The unordered pair {x,y}\{x,y\} and the singleton {x}={x,x}\{x\} = \{x,x\}).

[L4]

There is exactly one set with no elements, written \varnothing (There is exactly one set with no elements, written \varnothing).

Proof

technique · direct
1.1

Claim (i): zz \in \bigcup \varnothing requires a member ss of \varnothing with zsz \in s, and \varnothing has no members, so \bigcup \varnothing has no elements and is therefore \varnothing.

L1L4
1.2

Claim (ii): z{a}z \in \bigcup \{a\} requires a member ss of {a}\{a\} with zsz \in s, and the only member of {a}\{a\} is aa, so z{a}z \in \bigcup\{a\} holds exactly when zaz \in a.

L1L3
1.3

Claim (iii): aba \cup b is {a,b}\bigcup\{a,b\} by definition, and z{a,b}z \in \bigcup\{a,b\} requires a member ss of {a,b}\{a,b\} with zsz \in s; the members of {a,b}\{a,b\} are aa and bb, so zabz \in a \cup b holds exactly when zaz \in a or zbz \in b.

L1L3
1.4

Claim (iv): {a}\{a\} is nonempty since aa is a member, and z{a}z \in \bigcap\{a\} holds exactly when zz belongs to every member of {a}\{a\}, that is, exactly when zaz \in a.

L2L3
1.5

Claim (v): aba \cap b is {a,b}\bigcap\{a,b\} by definition, {a,b}\{a,b\} is nonempty, and z{a,b}z \in \bigcap\{a,b\} holds exactly when zz belongs to every member of {a,b}\{a,b\}; the members are aa and bb, so zabz \in a \cap b holds exactly when zaz \in a and zbz \in b.

L2L3
2.1

In each of the five claims the elements of the two sides have been shown to satisfy the same condition, so the sets are equal, which is the statement.

L5step 1.1step 1.2step 1.3step 1.4step 1.5
DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-08-06 (claude-opus-5)Open item page →

The difference aba \setminus b, the symmetric difference aba \triangle b, and the complement XaX \setminus a relative to a set XX

Definition

Let aa and bb be sets. The difference of aa and bb is

ab:={za:zb},a \setminus b := \{\, z \in a : z \notin b \,\},

a set by The Axiom Schema of Separation: for each formula φ\varphi, pˉxyz(zy(zxφ(z,pˉ)))\forall \bar p\,\forall x\,\exists y\,\forall z\,(z \in y \leftrightarrow (z \in x \wedge \varphi(z,\bar p))) and the only one with those elements by The Axiom of Extensionality: xy(z(zxzy)x=y)\forall x\,\forall y\,(\forall z\,(z \in x \leftrightarrow z \in y) \to x = y); so zabz \in a \setminus b holds exactly when zaz \in a and zbz \notin b. The symmetric difference is

ab:=(ab)(ba),a \triangle b := (a \setminus b) \cup (b \setminus a),

formed with the binary union of The union x\bigcup x of a set, and the binary union ab:={a,b}a \cup b := \bigcup \{a,b\}.

When aXa \subseteq X (Subset xyx \subseteq y, proper subset xyx \subsetneq y, and the separation notation {zx:φ(z)}\{\, z \in x : \varphi(z) \,\}), the difference XaX \setminus a is called the complement of aa relative to XX. The qualification is not optional: there is no set of all sets (There is no set UU with yUy \in U for every set yy), so an absolute complement, the class of all sets not in aa, is never a set.

Remarks

  • Difference needs no hypothesis. aba \setminus b is defined for all sets aa and bb, with no assumption that bab \subseteq a or that the two meet. When they are disjoint, ab=aa \setminus b = a.

  • Symmetric difference collects what lies in exactly one of the two sets. Unfolding the definition, zabz \in a \triangle b holds exactly when zz belongs to aa or to bb but not to both.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

For every set xx there is exactly one set whose elements are precisely the subsets of xx

Statement

For every set xx there is exactly one set pp such that, for every zz, zpz \in p holds if and only if zxz \subseteq x.

Facts & Assumptions

Proof

technique · direct
1.1

Power Set supplies a set yy that contains every zz all of whose elements belong to xx; by the meaning of inclusion, yy contains every zz with zxz \subseteq x.

L1L3
2.1

Apply Separation to yy with the formula φ(z,x):=t(tztx)\varphi(z,x) := \forall t\,(t \in z \to t \in x) and the parameter xx: the set p:={zy:zx}p := \{\, z \in y : z \subseteq x \,\} exists, and for every zz, zpz \in p holds if and only if zyz \in y and zxz \subseteq x.

L2L3step 1.1
3.1

If zxz \subseteq x then zyz \in y by step 1.1, so the condition "zyz \in y and zxz \subseteq x" reduces to zxz \subseteq x; hence zpz \in p holds if and only if zxz \subseteq x, which proves existence.

step 1.1step 2.1
4.1

If pp' also satisfies "zpz \in p' if and only if zxz \subseteq x" for every zz, then zpz \in p holds if and only if zpz \in p' for every zz, so p=pp = p'; existence and uniqueness together give the statement.

L4step 3.1
DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-08-06 (claude-opus-5)Open item page →

The power set P(x)={z:zx}\mathcal{P}(x) = \{\, z : z \subseteq x \,\}

Definition

Let xx be a set. By For every set xx there is exactly one set whose elements are precisely the subsets of xx there is exactly one set whose elements are precisely the subsets of xx (Subset xyx \subseteq y, proper subset xyx \subsetneq y, and the separation notation {zx:φ(z)}\{\, z \in x : \varphi(z) \,\}); it is the power set of xx, written P(x)\mathcal{P}(x). Thus zP(x)z \in \mathcal{P}(x) holds if and only if zxz \subseteq x, and in the class notation of the page,

P(x)={z:zx}.\mathcal{P}(x) = \{\, z : z \subseteq x \,\}.

Remarks

PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

P()={}\mathcal{P}(\varnothing) = \{\varnothing\}; P(a)\varnothing \in \mathcal{P}(a) and aP(a)a \in \mathcal{P}(a); aba \subseteq b if and only if P(a)P(b)\mathcal{P}(a) \subseteq \mathcal{P}(b); P(a)=a\bigcup \mathcal{P}(a) = a; P(a)=\bigcap \mathcal{P}(a) = \varnothing; and P(ab)=P(a)P(b)\mathcal{P}(a \cap b) = \mathcal{P}(a) \cap \mathcal{P}(b) while only P(a)P(b)P(ab)\mathcal{P}(a) \cup \mathcal{P}(b) \subseteq \mathcal{P}(a \cup b) holds

Statement

For all sets aa and bb:

  • (i) P()={}\mathcal{P}(\varnothing) = \{\varnothing\};
  • (ii) P(a)\varnothing \in \mathcal{P}(a) and aP(a)a \in \mathcal{P}(a);
  • (iii) aba \subseteq b if and only if P(a)P(b)\mathcal{P}(a) \subseteq \mathcal{P}(b);
  • (iv) P(a)=a\bigcup \mathcal{P}(a) = a;
  • (v) P(a)=\bigcap \mathcal{P}(a) = \varnothing;
  • (vi) P(ab)=P(a)P(b)\mathcal{P}(a \cap b) = \mathcal{P}(a) \cap \mathcal{P}(b);
  • (vii) P(a)P(b)P(ab)\mathcal{P}(a) \cup \mathcal{P}(b) \subseteq \mathcal{P}(a \cup b);
  • (viii) the inclusion in (vii) is an equality if and only if aba \subseteq b or bab \subseteq a.

Facts & Assumptions

Given: sets aa and bb.

[L1]

zP(x)z \in \mathcal{P}(x) holds if and only if zxz \subseteq x (The power set P(x)={z:zx}\mathcal{P}(x) = \{\, z : z \subseteq x \,\}).

[L6]

x\bigcup x is the set whose elements are exactly the elements of the elements of xx (The union x\bigcup x of a set, and the binary union ab:={a,b}a \cup b := \bigcup \{a,b\}).

[L7]

For xx \neq \varnothing, x\bigcap x is the set whose elements are exactly the sets belonging to every element of xx (The intersection x\bigcap x of a nonempty set, the binary intersection ab:={a,b}a \cap b := \bigcap\{a,b\}, and disjointness).

[L10]

There is exactly one set with no elements, written \varnothing (There is exactly one set with no elements, written \varnothing).

[L11]

{x}:={x,x}\{x\} := \{x,x\}, the singleton of xx, is the set whose only element is xx (The unordered pair {x,y}\{x,y\} and the singleton {x}={x,x}\{x\} = \{x,x\}).

Proof

technique · direct
1.1

Claim (i): zP()z \in \mathcal{P}(\varnothing) holds exactly when zz \subseteq \varnothing, that is, when every element of zz lies in \varnothing; since \varnothing has no elements this says zz has no elements, hence z=z = \varnothing. So the elements of P()\mathcal{P}(\varnothing) are exactly \varnothing, and P()={}\mathcal{P}(\varnothing) = \{\varnothing\}.

L1L2L10L11L12
1.2

Claim (ii): a\varnothing \subseteq a and aaa \subseteq a, so both \varnothing and aa are elements of P(a)\mathcal{P}(a).

L1L3L4
1.3

Claim (iii): if aba \subseteq b and zP(a)z \in \mathcal{P}(a), then zaz \subseteq a and aba \subseteq b give zbz \subseteq b, so zP(b)z \in \mathcal{P}(b); conversely if P(a)P(b)\mathcal{P}(a) \subseteq \mathcal{P}(b) then aP(a)a \in \mathcal{P}(a), since aaa \subseteq a, so aP(b)a \in \mathcal{P}(b), which says aba \subseteq b.

L1L2L4L5
1.4

Claim (iv): zP(a)z \in \bigcup\mathcal{P}(a) holds exactly when zz belongs to some element of P(a)\mathcal{P}(a), that is, to some ss with sas \subseteq a, and any such zz lies in aa; conversely if zaz \in a then zz lies in the element aa of P(a)\mathcal{P}(a). So the two sets have the same elements.

L1L2L4L6L12
1.5

Claim (v): P(a)\varnothing \in \mathcal{P}(a), so P(a)\mathcal{P}(a) \neq \varnothing and P(a)\bigcap\mathcal{P}(a) is defined; an element of it would have to belong to every element of P(a)\mathcal{P}(a), in particular to \varnothing, which has no elements. So P(a)\bigcap\mathcal{P}(a) has no elements and equals \varnothing.

L1L3L7L10
1.6

Claim (vi): zP(ab)z \in \mathcal{P}(a \cap b) holds exactly when every element of zz lies in aba \cap b, that is, lies in aa and in bb; this is exactly zaz \subseteq a together with zbz \subseteq b, that is, zP(a)z \in \mathcal{P}(a) and zP(b)z \in \mathcal{P}(b), which is zP(a)P(b)z \in \mathcal{P}(a) \cap \mathcal{P}(b).

L1L2L9L12
1.7

Claim (vii): if zP(a)P(b)z \in \mathcal{P}(a) \cup \mathcal{P}(b) then zaz \subseteq a or zbz \subseteq b; in either case every element of zz lies in aa or in bb, hence in aba \cup b, so zP(ab)z \in \mathcal{P}(a \cup b).

L1L2L8
2.1

Claim (viii): if aba \subseteq b then the elements of aba \cup b are exactly those of bb, so zabz \subseteq a \cup b is equivalent to zbz \subseteq b and P(ab)=P(b)P(a)P(b)\mathcal{P}(a \cup b) = \mathcal{P}(b) \subseteq \mathcal{P}(a) \cup \mathcal{P}(b), which with the reverse inclusion gives equality; the case bab \subseteq a is symmetric. Conversely, if equality holds then ababa \cup b \subseteq a \cup b puts aba \cup b in P(a)P(b)\mathcal{P}(a) \cup \mathcal{P}(b), so abaa \cup b \subseteq a or abba \cup b \subseteq b, and since babb \subseteq a \cup b and aaba \subseteq a \cup b these give bab \subseteq a and aba \subseteq b respectively.

L1L2L4L5L8step 1.7
3.1

Claims (i) to (viii) are established, which is the statement.

step 1.1step 1.2step 1.3step 1.4step 1.5step 1.6step 1.7step 2.1
PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

Union and intersection are commutative, associative, idempotent and absorptive; a=aa \cup \varnothing = a, a=a \cap \varnothing = \varnothing, a=aa \setminus \varnothing = a, aa=a \setminus a = \varnothing; and aba \subseteq b if and only if ab=ba \cup b = b, if and only if ab=aa \cap b = a, if and only if ab=a \setminus b = \varnothing

Statement

For all sets aa, bb and cc:

  • (i) ab=baa \cup b = b \cup a and ab=baa \cap b = b \cap a;
  • (ii) (ab)c=a(bc)(a \cup b) \cup c = a \cup (b \cup c) and (ab)c=a(bc)(a \cap b) \cap c = a \cap (b \cap c);
  • (iii) aa=aa \cup a = a and aa=aa \cap a = a;
  • (iv) a(ab)=aa \cup (a \cap b) = a and a(ab)=aa \cap (a \cup b) = a;
  • (v) a=aa \cup \varnothing = a, a=a \cap \varnothing = \varnothing, a=aa \setminus \varnothing = a and aa=a \setminus a = \varnothing;
  • (vi) aba \subseteq b, ab=ba \cup b = b, ab=aa \cap b = a and ab=a \setminus b = \varnothing are equivalent.

Facts & Assumptions

Given: sets aa, bb and cc.

[L3]

zabz \in a \setminus b holds exactly when zaz \in a and zbz \notin b (The difference aba \setminus b, the symmetric difference aba \triangle b, and the complement XaX \setminus a relative to a set XX).

[L6]

There is exactly one set with no elements, written \varnothing (There is exactly one set with no elements, written \varnothing).

Proof

technique · direct
1.1

Claim (i): zabz \in a \cup b holds exactly when zaz \in a or zbz \in b, which is the same condition as zbz \in b or zaz \in a, so aba \cup b and bab \cup a have the same elements; the same argument with "and" in place of "or" gives ab=baa \cap b = b \cap a.

L1L2L5
1.2

Claim (ii): z(ab)cz \in (a \cup b) \cup c holds exactly when zaz \in a or zbz \in b or zcz \in c, in either bracketing, so the two sides have the same elements; replacing "or" by "and" throughout gives the statement for intersection.

L1L2L5
1.3

Claim (iii): zaaz \in a \cup a holds exactly when zaz \in a or zaz \in a, and zaaz \in a \cap a exactly when zaz \in a and zaz \in a; both conditions are zaz \in a.

L1L2L5
1.4

Claim (iv): za(ab)z \in a \cup (a \cap b) holds exactly when zaz \in a, or zaz \in a and zbz \in b, which holds exactly when zaz \in a; and za(ab)z \in a \cap (a \cup b) exactly when zaz \in a, and zaz \in a or zbz \in b, which again holds exactly when zaz \in a.

L1L2L5
1.5

Claim (v): zaz \in a \cup \varnothing exactly when zaz \in a or zz \in \varnothing, and the second alternative never holds, so this is zaz \in a; zaz \in a \cap \varnothing requires zz \in \varnothing, so aa \cap \varnothing has no elements; zaz \in a \setminus \varnothing exactly when zaz \in a and zz \notin \varnothing, and the second condition always holds; zaaz \in a \setminus a requires zaz \in a and zaz \notin a, so aaa \setminus a has no elements.

L1L2L3L5L6
1.6

Claim (vi), that aba \subseteq b and ab=ba \cup b = b are equivalent: if aba \subseteq b then zabz \in a \cup b, which says zaz \in a or zbz \in b, holds exactly when zbz \in b, so ab=ba \cup b = b; conversely if ab=ba \cup b = b and zaz \in a, then zab=bz \in a \cup b = b.

L1L4L5
1.7

Claim (vi), that aba \subseteq b and ab=aa \cap b = a are equivalent: if aba \subseteq b then zabz \in a \cap b, which says zaz \in a and zbz \in b, holds exactly when zaz \in a, so ab=aa \cap b = a; conversely if ab=aa \cap b = a and zaz \in a, then zabz \in a \cap b, so zbz \in b.

L2L4L5
1.8

Claim (vi), that aba \subseteq b and ab=a \setminus b = \varnothing are equivalent: if aba \subseteq b then no zz satisfies zaz \in a and zbz \notin b, so aba \setminus b has no elements; conversely if ab=a \setminus b = \varnothing and zaz \in a, then zbz \notin b is impossible, so zbz \in b.

L3L4L6
2.1

Claims (i) to (vi) are established, and the four conditions of (vi) are equivalent because each of the last three was shown equivalent to the first.

step 1.1step 1.2step 1.3step 1.4step 1.5step 1.6step 1.7step 1.8
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

a(bc)=(ab)(ac)a \cap (b \cup c) = (a \cap b) \cup (a \cap c) and a(bc)=(ab)(ac)a \cup (b \cap c) = (a \cup b) \cap (a \cup c); and for a nonempty set FF, aF={ab:bF}a \cap \bigcup F = \bigcup \{\, a \cap b : b \in F \,\} and aF={ab:bF}a \cup \bigcap F = \bigcap \{\, a \cup b : b \in F \,\}

Statement

For all sets aa, bb and cc,

a(bc)=(ab)(ac),a(bc)=(ab)(ac).a \cap (b \cup c) = (a \cap b) \cup (a \cap c), \qquad a \cup (b \cap c) = (a \cup b) \cap (a \cup c).

Let FF be a set with FF \neq \varnothing. Then {ab:bF}\{\, a \cap b : b \in F \,\} and {ab:bF}\{\, a \cup b : b \in F \,\} are nonempty sets, and

aF={ab:bF},aF={ab:bF}.a \cap \bigcup F = \bigcup \{\, a \cap b : b \in F \,\}, \qquad a \cup \bigcap F = \bigcap \{\, a \cup b : b \in F \,\}.

Facts & Assumptions

Given: sets aa, bb, cc and a set FF with FF \neq \varnothing where the family forms are concerned.

[L3]

x\bigcup x is the set whose elements are exactly the elements of the elements of xx (The union x\bigcup x of a set, and the binary union ab:={a,b}a \cup b := \bigcup \{a,b\}).

[L4]

For xx \neq \varnothing, x\bigcap x is the set whose elements are exactly the sets belonging to every element of xx (The intersection x\bigcap x of a nonempty set, the binary intersection ab:={a,b}a \cap b := \bigcap\{a,b\}, and disjointness).

[L6]

zP(x)z \in \mathcal{P}(x) holds if and only if zxz \subseteq x (The power set P(x)={z:zx}\mathcal{P}(x) = \{\, z : z \subseteq x \,\}).

Proof

technique · direct
1.1

Every aba \cap b is a subset of aa, so the class {ab:bF}\{\, a \cap b : b \in F \,\} is the separated set {wP(a):b(bFw=ab)}\{\, w \in \mathcal{P}(a) : \exists b\,(b \in F \wedge w = a \cap b) \,\}, hence a set; and it has a member because FF does.

L2L5L6L7
1.2

Every aba \cup b with bFb \in F is a subset of aFa \cup \bigcup F, since an element of bb lies in F\bigcup F; so {ab:bF}\{\, a \cup b : b \in F \,\} is the separated set {wP(aF):b(bFw=ab)}\{\, w \in \mathcal{P}(a \cup \bigcup F) : \exists b\,(b \in F \wedge w = a \cup b) \,\}, hence a set, and it has a member because FF does.

L1L3L5L6L7
1.3

za(bc)z \in a \cap (b \cup c) holds exactly when zaz \in a, and zbz \in b or zcz \in c; that is exactly when zaz \in a and zbz \in b, or zaz \in a and zcz \in c, which is z(ab)(ac)z \in (a \cap b) \cup (a \cap c).

L1L2L8
1.4

za(bc)z \in a \cup (b \cap c) holds exactly when zaz \in a, or zbz \in b and zcz \in c; that is exactly when zaz \in a or zbz \in b, and zaz \in a or zcz \in c, which is z(ab)(ac)z \in (a \cup b) \cap (a \cup c).

L1L2L8
2.1

zaFz \in a \cap \bigcup F holds exactly when zaz \in a and zsz \in s for some sFs \in F, that is, exactly when zasz \in a \cap s for some sFs \in F; the sets asa \cap s with sFs \in F are precisely the elements of {ab:bF}\{\, a \cap b : b \in F \,\}, so this holds exactly when zz lies in some element of that set, which is z{ab:bF}z \in \bigcup \{\, a \cap b : b \in F \,\}.

L2L3L8step 1.1
2.2

z{ab:bF}z \in \bigcap \{\, a \cup b : b \in F \,\} holds exactly when, for every bFb \in F, zaz \in a or zbz \in b. If zaFz \in a \cup \bigcap F then either zaz \in a, and the condition holds outright, or zz lies in every element of FF, and it holds again. Conversely, if the condition holds and zaz \notin a, then zbz \in b for every bFb \in F, so zFz \in \bigcap F; hence zaFz \in a \cup \bigcap F in both cases.

L1L4L8step 1.2
3.1

The two finite laws and the two family laws are established, which is the statement.

step 1.3step 1.4step 2.1step 2.2
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

X(ab)=(Xa)(Xb)X \setminus (a \cup b) = (X \setminus a) \cap (X \setminus b) and X(ab)=(Xa)(Xb)X \setminus (a \cap b) = (X \setminus a) \cup (X \setminus b); and for a nonempty set FF, XF={Xa:aF}X \setminus \bigcup F = \bigcap \{\, X \setminus a : a \in F \,\} and XF={Xa:aF}X \setminus \bigcap F = \bigcup \{\, X \setminus a : a \in F \,\}

Statement

For all sets XX, aa and bb,

X(ab)=(Xa)(Xb),X(ab)=(Xa)(Xb).X \setminus (a \cup b) = (X \setminus a) \cap (X \setminus b), \qquad X \setminus (a \cap b) = (X \setminus a) \cup (X \setminus b).

Let FF be a set with FF \neq \varnothing. Then {Xa:aF}\{\, X \setminus a : a \in F \,\} is a nonempty set and

XF={Xa:aF},XF={Xa:aF}.X \setminus \bigcup F = \bigcap \{\, X \setminus a : a \in F \,\}, \qquad X \setminus \bigcap F = \bigcup \{\, X \setminus a : a \in F \,\}.

Facts & Assumptions

Given: sets XX, aa, bb, and a set FF with FF \neq \varnothing where the family forms are concerned.

[L1]

zabz \in a \setminus b holds exactly when zaz \in a and zbz \notin b (The difference aba \setminus b, the symmetric difference aba \triangle b, and the complement XaX \setminus a relative to a set XX).

[L4]

x\bigcup x is the set whose elements are exactly the elements of the elements of xx (The union x\bigcup x of a set, and the binary union ab:={a,b}a \cup b := \bigcup \{a,b\}).

[L5]

For xx \neq \varnothing, x\bigcap x is the set whose elements are exactly the sets belonging to every element of xx (The intersection x\bigcap x of a nonempty set, the binary intersection ab:={a,b}a \cap b := \bigcap\{a,b\}, and disjointness).

[L7]

zP(x)z \in \mathcal{P}(x) holds if and only if zxz \subseteq x (The power set P(x)={z:zx}\mathcal{P}(x) = \{\, z : z \subseteq x \,\}).

Proof

technique · direct
1.1

Every XaX \setminus a is a subset of XX, so the class {Xa:aF}\{\, X \setminus a : a \in F \,\} is the separated set {wP(X):a(aFw=Xa)}\{\, w \in \mathcal{P}(X) : \exists a\,(a \in F \wedge w = X \setminus a) \,\}, hence a set; and it has a member because FF does.

L1L6L7L8
1.2

zX(ab)z \in X \setminus (a \cup b) holds exactly when zXz \in X and neither zaz \in a nor zbz \in b, that is, exactly when zXz \in X and zaz \notin a, and zXz \in X and zbz \notin b; this is z(Xa)(Xb)z \in (X \setminus a) \cap (X \setminus b).

L1L2L3L9
1.3

zX(ab)z \in X \setminus (a \cap b) holds exactly when zXz \in X and not both zaz \in a and zbz \in b, that is, exactly when zXz \in X and zaz \notin a, or zXz \in X and zbz \notin b; this is z(Xa)(Xb)z \in (X \setminus a) \cup (X \setminus b).

L1L2L3L9
2.1

zXFz \in X \setminus \bigcup F holds exactly when zXz \in X and no member ss of FF has zsz \in s, that is, exactly when zXaz \in X \setminus a for every aFa \in F; the sets XaX \setminus a with aFa \in F are precisely the elements of {Xa:aF}\{\, X \setminus a : a \in F \,\}, which is nonempty, so this is z{Xa:aF}z \in \bigcap \{\, X \setminus a : a \in F \,\}.

L1L4L5L9step 1.1
2.2

zXFz \in X \setminus \bigcap F holds exactly when zXz \in X and zz fails to lie in some member aa of FF, that is, exactly when zXaz \in X \setminus a for some aFa \in F; this is z{Xa:aF}z \in \bigcup \{\, X \setminus a : a \in F \,\}.

L1L4L5L9step 1.1
3.1

The two finite laws and the two family laws are established, which is the statement.

step 1.2step 1.3step 2.1step 2.2
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

If aba \subseteq b then ab\bigcup a \subseteq \bigcup b; if in addition aa \neq \varnothing then ba\bigcap b \subseteq \bigcap a; and cac \subseteq \bigcup a for every cac \in a, while ac\bigcap a \subseteq c for every cac \in a

Statement

Let aa and bb be sets.

  • (i) If aba \subseteq b then ab\bigcup a \subseteq \bigcup b.
  • (ii) If aba \subseteq b and aa \neq \varnothing, then bb \neq \varnothing, both intersections are defined, and ba\bigcap b \subseteq \bigcap a.
  • (iii) cac \subseteq \bigcup a for every cac \in a.
  • (iv) ac\bigcap a \subseteq c for every cac \in a.

Facts & Assumptions

Given: sets aa and bb.

[L1]

x\bigcup x is the set whose elements are exactly the elements of the elements of xx (The union x\bigcup x of a set, and the binary union ab:={a,b}a \cup b := \bigcup \{a,b\}).

[L2]

For xx \neq \varnothing, x\bigcap x is the set whose elements are exactly the sets belonging to every element of xx (The intersection x\bigcap x of a nonempty set, the binary intersection ab:={a,b}a \cap b := \bigcap\{a,b\}, and disjointness).

[L4]

There is exactly one set with no elements, written \varnothing (There is exactly one set with no elements, written \varnothing).

Proof

technique · direct
1.1

Claim (i): assume aba \subseteq b and let zaz \in \bigcup a; then zsz \in s for some sas \in a, and sbs \in b because aba \subseteq b, so zbz \in \bigcup b.

L1L3
1.2

Claim (ii): assume aba \subseteq b and aa \neq \varnothing. Then aa has a member, which is also a member of bb, so bb \neq \varnothing and both intersections are defined. If zbz \in \bigcap b then zz lies in every member of bb, hence in every member of aa, so zaz \in \bigcap a.

L2L3L4
1.3

Claim (iii): let cac \in a and zcz \in c; then zz lies in a member of aa, so zaz \in \bigcup a, and therefore cac \subseteq \bigcup a.

L1L3
1.4

Claim (iv): let cac \in a; then aa \neq \varnothing, so a\bigcap a is defined, and every zaz \in \bigcap a lies in every member of aa, in particular in cc; therefore ac\bigcap a \subseteq c.

L2L3L4
2.1

Claims (i) to (iv) are established, which is the statement.

step 1.1step 1.2step 1.3step 1.4
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

There is no RR with xRxxx \in R \leftrightarrow x \notin x for every xx

Statement

There is no set RR such that, for every set xx,

xRxx.x \in R \leftrightarrow x \notin x .

Facts & Assumptions

Given: the language of set theory, in which xRx \in R and xxx \notin x are formulas (The first-order language of set theory: \in, ==, formulas with parameters, and class abbreviations).

Proof

technique · contradiction
1.1

Suppose there is a set RR such that xRx \in R holds if and only if xxx \notin x, for every set xx.

assume-contra
2.1

The hypothesis holds for every set xx, and RR is a set, so it holds for x:=Rx := R: RRR \in R if and only if RRR \notin R.

step 1.1
3.1

A statement equivalent to its own negation is contradictory: if RRR \in R then RRR \notin R, and if RRR \notin R then RRR \in R, so each alternative refutes itself. There is therefore no such RR.

step 2.1discharge-contradiction

Remarks

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

There is no set UU with yUy \in U for every set yy

Statement

There is no set UU such that yUy \in U for every set yy.

Facts & Assumptions

Given: nothing beyond the axioms cited below.

[L2]

The separated set is written {zx:φ(z,pˉ)}\{\, z \in x : \varphi(z,\bar p) \,\} (Subset xyx \subseteq y, proper subset xyx \subsetneq y, and the separation notation {zx:φ(z)}\{\, z \in x : \varphi(z) \,\}).

[L3]

There is no set RR such that, for every set xx, xRx \in R holds if and only if xxx \notin x (There is no RR with xRxxx \in R \leftrightarrow x \notin x for every xx).

Proof

technique · contradiction
1.1

Suppose UU is a set with yUy \in U for every set yy.

assume-contra
2.1

Separation applied to UU with the formula φ(x):=xx\varphi(x) := x \notin x gives the set R:={xU:xx}R := \{\, x \in U : x \notin x \,\}, whose elements are exactly the xUx \in U with xxx \notin x.

L1L2step 1.1
3.1

Every set xx satisfies xUx \in U, so for every set xx the condition "xUx \in U and xxx \notin x" reduces to xxx \notin x; hence xRx \in R holds if and only if xxx \notin x, for every set xx.

step 1.1step 2.1
4.1

Step 3.1 produces exactly the set that cannot exist, so the supposition is untenable and no such UU exists.

L3step 3.1discharge-contradiction

Remarks

CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

There is no set yy with xys(sxs)x \in y \leftrightarrow \forall s\,(s \in \varnothing \to x \in s), so \bigcap \varnothing is undefined

Statement

There is no set yy such that, for every set xx,

xys(sxs).x \in y \leftrightarrow \forall s\,(s \in \varnothing \to x \in s).

The defining condition for x\bigcap x therefore determines no set when x=x = \varnothing, which is why \bigcap \varnothing is left undefined.

Facts & Assumptions

Given: nothing beyond the results cited below.

[L1]

There is exactly one set with no elements, written \varnothing (There is exactly one set with no elements, written \varnothing).

[L2]

There is no set UU such that yUy \in U for every set yy (There is no set UU with yUy \in U for every set yy).

[L3]

For xx \neq \varnothing, x\bigcap x is the set whose elements are exactly the sets belonging to every element of xx (The intersection x\bigcap x of a nonempty set, the binary intersection ab:={a,b}a \cap b := \bigcap\{a,b\}, and disjointness).

Proof

technique · contradiction
1.1

Suppose there is a set yy such that, for every set xx, xyx \in y holds if and only if xx belongs to every element of \varnothing.

assume-contra
1.2

\varnothing has no elements, so for every set xx the condition "xx belongs to every element of \varnothing" is satisfied vacuously.

L1
2.1

Combining, every set xx satisfies xyx \in y.

step 1.1step 1.2
3.1

A set with every set as an element does not exist, so the supposition fails; the condition defining \bigcap therefore determines no set at \varnothing, and \bigcap \varnothing is undefined.

L2L3step 2.1discharge-contradiction
RemarkRemark: AI-adaptedProof: Not applicableverified 2026-08-06 (claude-opus-5)Open item page →

Separation and Replacement build subsets of sets already in hand, which is exactly what blocks Russell's construction

Remark

An unrestricted comprehension principle would assert, for every formula φ(x)\varphi(x) of the language (The first-order language of set theory: \in, ==, formulas with parameters, and class abbreviations), that yx(xyφ(x))\exists y\,\forall x\,(x \in y \leftrightarrow \varphi(x)). Taking φ(x):=xx\varphi(x) := x \notin x turns that assertion into precisely the set There is no RR with xRxxx \in R \leftrightarrow x \notin x for every xx shows cannot exist, so the principle is refutable in plain first-order logic; no appeal to any axiom below is needed to reject it.

The Axiom Schema of Separation: for each formula φ\varphi, pˉxyz(zy(zxφ(z,pˉ)))\forall \bar p\,\forall x\,\exists y\,\forall z\,(z \in y \leftrightarrow (z \in x \wedge \varphi(z,\bar p))) asserts something weaker in a specific way: it does not produce a set from a formula alone, but only from a formula together with a set xx already in hand, and the set it produces is a subset of that xx. Running Russell's argument against it therefore yields no contradiction but a theorem. Given any set xx, the separated set r:={zx:zz}r := \{\, z \in x : z \notin z \,\} exists; asking whether rrr \in r shows that rxr \in x is impossible, since rxr \in x would give rrrrr \in r \leftrightarrow r \notin r. So every set xx has a subset that is not one of its elements, and no set contains every set, which is There is no set UU with yUy \in U for every set yy.

The same restriction is what makes The Axiom Schema of Replacement: for each formula φ\varphi, if φ\varphi defines a class function on AA then its image on AA is a set safe. It does not assert that an arbitrary class is a set either: its hypothesis is that a formula behaves like a function on a set AA already in hand, and its conclusion is about the image of that particular set. Both schemas therefore build only from material already given, and neither can be turned on the universe at large.

What is given up is small and is worth naming exactly: from these axioms alone nothing whatever can be constructed, which is why The Axiom of Infinity: there is a set containing a set with no elements and closed under yy{y}y \mapsto y \cup \{y\} is assumed outright rather than derived, and why There is exactly one set with no elements, written \varnothing needs the logical fact that the domain of discourse is nonempty before Separation has anything to act on.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

Under Foundation, xxx \notin x for every set xx, there are no sets with xyxx \in y \in x, and there are no sets with xyzxx \in y \in z \in x

Statement

Assume the Axiom of Foundation. Then:

  • (i) xxx \notin x for every set xx;
  • (ii) there are no sets xx and yy with xyx \in y and yxy \in x;
  • (iii) there are no sets xx, yy and zz with xyx \in y, yzy \in z and zxz \in x.

Cycles of length greater than three are not treated here: stating "there is no membership cycle of any finite length" needs a finite sequence, that is a function on a natural number, and neither notion is available at this point in the reading order.

Facts & Assumptions

Proof

technique · contradiction
1.1

Suppose the statement fails, so that at least one of the following holds: there is a set xx with xxx \in x; there are sets xx and yy with xyx \in y and yxy \in x; there are sets xx, yy and zz with xyx \in y, yzy \in z and zxz \in x.

assume-contra
2.1

Suppose xxx \in x and put S:={x}S := \{x\}. Its only member is xx, so SS has a member and Foundation applies; the member it supplies must be xx, so xx shares no member with SS. But xxx \in x and xSx \in S, so xx is a member of both.

L1L2step 1.1
2.2

Suppose xyx \in y and yxy \in x, and put S:={x,y}S := \{x,y\}. Foundation supplies a member of SS sharing no member with SS, and that member is xx or yy. If it is xx, then yxy \in x and ySy \in S make yy a member of both; if it is yy, then xyx \in y and xSx \in S make xx a member of both.

L1L2step 1.1
2.3

Suppose xyx \in y, yzy \in z and zxz \in x, and put S:={x,y}{z}S := \{x,y\} \cup \{z\}, whose members are exactly xx, yy and zz. Foundation supplies a member of SS sharing no member with SS. If it is xx, then zxz \in x and zSz \in S; if it is yy, then xyx \in y and xSx \in S; if it is zz, then yzy \in z and ySy \in S. In each case that member shares a member with SS.

L1L2L3L4step 1.1
3.1

Each of the three suppositions contradicts the member Foundation supplies, so none of them holds: xxx \notin x for every xx, no sets satisfy xyxx \in y \in x, and no sets satisfy xyzxx \in y \in z \in x.

step 2.1step 2.2step 2.3discharge-contradiction
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

If x{x}=y{y}x \cup \{x\} = y \cup \{y\} then x=yx = y

Statement

For all sets xx and yy, if x{x}=y{y}x \cup \{x\} = y \cup \{y\} then x=yx = y.

Facts & Assumptions

Given: sets xx and yy with x{x}=y{y}x \cup \{x\} = y \cup \{y\}; the Axiom of Foundation, through the result cited as [L3].

[L2]

{x}:={x,x}\{x\} := \{x,x\}, the singleton of xx, is the set whose only element is xx (The unordered pair {x,y}\{x,y\} and the singleton {x}={x,x}\{x\} = \{x,x\}).

Proof

technique · direct
1.1

x{x}x \in \{x\}, so xx{x}x \in x \cup \{x\}; likewise yy{y}y \in y \cup \{y\}.

L1L2L4
2.1

By hypothesis the two sets are equal, so xy{y}x \in y \cup \{y\} and yx{x}y \in x \cup \{x\}; hence xyx \in y or x=yx = y, and yxy \in x or y=xy = x.

L1L2givenstep 1.1
3.1

If xyx \neq y then both alternatives x=yx = y and y=xy = x fail, leaving xyx \in y and yxy \in x, which is impossible; therefore x=yx = y.

L3step 2.1

Remarks

  • The general statement, and its special case for ω\omega. This holds for all sets and is proved from Foundation. The corresponding clause for the natural numbers, that the successor is injective on ω\omega, is part (P2) of The von Neumann naturals form a Peano system , which proves it by a different route that does not use Foundation.
DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-08-06 (claude-opus-5)Open item page →

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

Definition

For sets aa and bb, the ordered pair (a,b)(a,b) is the set

(a,b):={{a},{a,b}},(a,b) := \{\{a\},\{a,b\}\},

formed from the unordered pairs and singletons of The unordered pair {x,y}\{x,y\} and the singleton {x}={x,x}\{x\} = \{x,x\}; three applications of The Axiom of Pairing: xyzt(tz(t=xt=y))\forall x\,\forall y\,\exists z\,\forall t\,(t \in z \leftrightarrow (t = x \vee t = y)) produce it. The first coordinate is aa and the second is bb.

When a=ba = b the two members coincide, since {a,b}={a,a}={a}\{a,b\} = \{a,a\} = \{a\}, and the pair degenerates to (a,a)={{a}}(a,a) = \{\{a\}\}.

Remarks

  • Why this set and not another. An ordered pair is required to satisfy one property, that (a,b)=(c,d)(a,b) = (c,d) exactly when a=ca = c and b=db = d; that is (a,b)=(c,d)(a,b) = (c,d) if and only if a=ca = c and b=db = d, and it is the only thing any later construction uses. Other definitions with the same property exist, and nothing below distinguishes them from this one.

  • The degenerate case is where a careless proof fails. An argument that treats {{a},{a,b}}\{\{a\},\{a,b\}\} as a set with two distinct members breaks at a=ba = b, and that case has to be handled separately in the proof of the characterising property.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

(a,b)=(c,d)(a,b) = (c,d) if and only if a=ca = c and b=db = d

Statement

For all sets aa, bb, cc, dd: (a,b)=(c,d)(a,b) = (c,d) if and only if a=ca = c and b=db = d.

Facts & Assumptions

Given: sets aa, bb, cc, dd.

[L1]

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

[L2]

{x,y}\{x,y\} is the set whose elements are exactly xx and yy, and {x}:={x,x}\{x\} := \{x,x\} (The unordered pair {x,y}\{x,y\} and the singleton {x}={x,x}\{x\} = \{x,x\}).

[L3]

{a,b}={c,d}  ((a=cb=d)(a=db=c))\{a,b\} = \{c,d\} \ \leftrightarrow\ \bigl((a = c \wedge b = d) \vee (a = d \wedge b = c)\bigr) ({a,b}={c,d}\{a,b\} = \{c,d\} if and only if (a=ca = c and b=db = d) or (a=da = d and b=cb = c); in particular {a}={c,d}\{a\} = \{c,d\} if and only if a=c=da = c = d).

Proof

technique · cases
1.1

Right to left: if a=ca = c and b=db = d then {a}={c}\{a\} = \{c\} and {a,b}={c,d}\{a,b\} = \{c,d\}, so {{a},{a,b}}\{\{a\},\{a,b\}\} and {{c},{c,d}}\{\{c\},\{c,d\}\} are the same set, that is (a,b)=(c,d)(a,b) = (c,d).

L1L2
1.2

Left to right, setting up: assume {{a},{a,b}}={{c},{c,d}}\{\{a\},\{a,b\}\} = \{\{c\},\{c,d\}\}. Applying the equality criterion for unordered pairs to these two sets, either {a}={c}\{a\} = \{c\} and {a,b}={c,d}\{a,b\} = \{c,d\}, or {a}={c,d}\{a\} = \{c,d\} and {a,b}={c}\{a,b\} = \{c\}.

L1L3
2.1

In the first alternative, {a}={c}\{a\} = \{c\} reads {a,a}={c,c}\{a,a\} = \{c,c\} and gives a=ca = c. Then {a,b}={c,d}\{a,b\} = \{c,d\} gives a=ca = c and b=db = d, or a=da = d and b=cb = c; in the second of these b=c=a=db = c = a = d, so b=db = d holds as well. Either way a=ca = c and b=db = d.

assume-case firststep 1.2L2L3
2.2

In the second alternative, {a,b}={c}\{a,b\} = \{c\} reads {a,b}={c,c}\{a,b\} = \{c,c\} and gives a=ca = c and b=cb = c, while {a}={c,d}\{a\} = \{c,d\} reads {a,a}={c,d}\{a,a\} = \{c,d\} and gives a=ca = c and a=da = d. Hence a=ca = c and b=c=a=db = c = a = d.

assume-case secondstep 1.2L2L3
3.1

The two alternatives supplied by step 1.2 are exhaustive, so the left-to-right implication holds.

step 2.1step 2.2cases-exhaustive
4.1

Both implications hold, which is the statement.

step 1.1step 3.1
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

If aAa \in A and bBb \in B then (a,b)P(P(AB))(a,b) \in \mathcal{P}(\mathcal{P}(A \cup B))

Statement

Let AA and BB be sets. If aAa \in A and bBb \in B, then (a,b)P(P(AB))(a,b) \in \mathcal{P}(\mathcal{P}(A \cup B)).

Facts & Assumptions

Given: sets AA and BB, and elements aAa \in A and bBb \in B.

[L1]

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

[L2]

{x,y}\{x,y\} is the set whose elements are exactly xx and yy, and {x}:={x,x}\{x\} := \{x,x\} (The unordered pair {x,y}\{x,y\} and the singleton {x}={x,x}\{x\} = \{x,x\}).

[L4]

zP(x)z \in \mathcal{P}(x) holds if and only if zxz \subseteq x (The power set P(x)={z:zx}\mathcal{P}(x) = \{\, z : z \subseteq x \,\}).

Proof

technique · direct
1.1

aAa \in A gives aABa \in A \cup B, and bBb \in B gives bABb \in A \cup B.

L3L6given
2.1

The elements of {a}\{a\} are aa alone and the elements of {a,b}\{a,b\} are aa and bb, so both sets are included in ABA \cup B and are therefore elements of P(AB)\mathcal{P}(A \cup B).

L2L4L5step 1.1
3.1

The elements of {{a},{a,b}}\{\{a\},\{a,b\}\} are {a}\{a\} and {a,b}\{a,b\}, so that set is included in P(AB)\mathcal{P}(A \cup B) and is therefore an element of P(P(AB))\mathcal{P}(\mathcal{P}(A \cup B)); and that set is (a,b)(a,b).

L1L2L4L5step 2.1
DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-08-06 (claude-opus-5)Open item page →

The Cartesian product A×B:={zP(P(AB)):aA bB 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) \,\}

Definition

Let AA and BB be sets. By If aAa \in A and bBb \in B then (a,b)P(P(AB))(a,b) \in \mathcal{P}(\mathcal{P}(A \cup B)) every ordered pair (a,b)(a,b) with aAa \in A and bBb \in B is an element of P(P(AB))\mathcal{P}(\mathcal{P}(A \cup B)), so The Axiom Schema of Separation: for each formula φ\varphi, pˉxyz(zy(zxφ(z,pˉ)))\forall \bar p\,\forall x\,\exists y\,\forall z\,(z \in y \leftrightarrow (z \in x \wedge \varphi(z,\bar p))) applied inside that set produces the Cartesian product

A×B:={zP(P(AB)):aA bB 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) \,\}.

Its elements are exactly the ordered pairs (a,b)(a,b) (The Kuratowski ordered pair (a,b):={{a},{a,b}}(a,b) := \{\{a\},\{a,b\}\}) with aAa \in A and bBb \in B: no such pair is lost, because each such pair lies in the ambient set being separated. Thus zA×Bz \in A \times B holds if and only if z=(a,b)z = (a,b) for some aAa \in A and some bBb \in B.

Remarks

PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

A×(BC)=(A×B)(A×C)A \times (B \cup C) = (A \times B) \cup (A \times C), A×(BC)=(A×B)(A×C)A \times (B \cap C) = (A \times B) \cap (A \times C), A×(BC)=(A×B)(A×C)A \times (B \setminus C) = (A \times B) \setminus (A \times C), (AB)×(CD)=(A×C)(B×D)(A \cap B) \times (C \cap D) = (A \times C) \cap (B \times D); A×B=A \times B = \varnothing if and only if A=A = \varnothing or B=B = \varnothing; and for nonempty AA and BB, A×BC×DA \times B \subseteq C \times D if and only if ACA \subseteq C and BDB \subseteq D

Statement

For all sets AA, BB, CC, DD:

  • (i) A×(BC)=(A×B)(A×C)A \times (B \cup C) = (A \times B) \cup (A \times C);
  • (ii) A×(BC)=(A×B)(A×C)A \times (B \cap C) = (A \times B) \cap (A \times C);
  • (iii) A×(BC)=(A×B)(A×C)A \times (B \setminus C) = (A \times B) \setminus (A \times C);
  • (iv) (AB)×(CD)=(A×C)(B×D)(A \cap B) \times (C \cap D) = (A \times C) \cap (B \times D);
  • (v) A×B=A \times B = \varnothing if and only if A=A = \varnothing or B=B = \varnothing;
  • (vi) if AA \neq \varnothing and BB \neq \varnothing, then A×BC×DA \times B \subseteq C \times D if and only if ACA \subseteq C and BDB \subseteq D.

Facts & Assumptions

Given: sets AA, BB, CC, DD.

[L2]

(a,b)=(c,d)(a,b) = (c,d) if and only if a=ca = c and b=db = d ((a,b)=(c,d)(a,b) = (c,d) if and only if a=ca = c and b=db = d).

[L5]

zabz \in a \setminus b holds exactly when zaz \in a and zbz \notin b (The difference aba \setminus b, the symmetric difference aba \triangle b, and the complement XaX \setminus a relative to a set XX).

[L7]

There is exactly one set with no elements, written \varnothing (There is exactly one set with no elements, written \varnothing).

[L9]

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

[L10]

ab:={a,b}a \cup b := \bigcup\{a,b\}, and x\bigcup x is the set whose elements are exactly the elements of the elements of xx (The union x\bigcup x of a set, and the binary union ab:={a,b}a \cup b := \bigcup \{a,b\}).

[L11]

For xx \neq \varnothing, x\bigcap x is the set whose elements are exactly the sets belonging to every element of xx (The intersection x\bigcap x of a nonempty set, the binary intersection ab:={a,b}a \cap b := \bigcap\{a,b\}, and disjointness).

Proof

technique · direct
1.1

Membership criterion: for all sets aa and bb, (a,b)A×B(a,b) \in A \times B holds if and only if aAa \in A and bBb \in B. Indeed, an element of A×BA \times B is a pair (a,b)(a',b') with aAa' \in A and bBb' \in B, and (a,b)=(a,b)(a,b) = (a',b') forces a=aa = a' and b=bb = b'; the converse is immediate from the description of A×BA \times B. Every element of a product is a pair, so it suffices in each identity below to compare pairs.

L1L2L9
2.1

Claim (i): (a,t)A×(BC)(a,t) \in A \times (B \cup C) holds exactly when aAa \in A and tBt \in B or tCt \in C, that is, exactly when (a,t)A×B(a,t) \in A \times B or (a,t)A×C(a,t) \in A \times C.

L3L8L10step 1.1
2.2

Claim (ii): (a,t)A×(BC)(a,t) \in A \times (B \cap C) holds exactly when aAa \in A, tBt \in B and tCt \in C, that is, exactly when (a,t)A×B(a,t) \in A \times B and (a,t)A×C(a,t) \in A \times C.

L4L8L11step 1.1
2.3

Claim (iii): (a,t)A×(BC)(a,t) \in A \times (B \setminus C) holds exactly when aAa \in A, tBt \in B and tCt \notin C. On the other side, (a,t)(A×B)(A×C)(a,t) \in (A \times B) \setminus (A \times C) holds exactly when aAa \in A, tBt \in B, and it is not the case that aAa \in A and tCt \in C; given aAa \in A, that last condition is tCt \notin C.

L5L8step 1.1
2.4

Claim (iv): (u,v)(AB)×(CD)(u,v) \in (A \cap B) \times (C \cap D) holds exactly when uAu \in A, uBu \in B, vCv \in C and vDv \in D, that is, exactly when (u,v)A×C(u,v) \in A \times C and (u,v)B×D(u,v) \in B \times D.

L4L8L11step 1.1
2.5

Claim (v): if A=A = \varnothing or B=B = \varnothing then no pair satisfies the membership criterion, so A×BA \times B has no elements and equals \varnothing; conversely if both are nonempty, fix aAa \in A and bBb \in B, and then (a,b)A×B(a,b) \in A \times B.

L7step 1.1
2.6

Claim (vi): assume AA \neq \varnothing and BB \neq \varnothing. If A×BC×DA \times B \subseteq C \times D, fix b0Bb_{0} \in B; for any aAa \in A the pair (a,b0)(a,b_{0}) lies in A×BA \times B, hence in C×DC \times D, so aCa \in C, and ACA \subseteq C follows; fixing a0Aa_{0} \in A and running the same argument on the second coordinate gives BDB \subseteq D. Conversely, if ACA \subseteq C and BDB \subseteq D, then any (a,b)A×B(a,b) \in A \times B has aCa \in C and bDb \in D, so it lies in C×DC \times D.

L6L7step 1.1
3.1

Claims (i) to (vi) are established, which is the statement.

step 2.1step 2.2step 2.3step 2.4step 2.5step 2.6
DefinitionDefinition: AI-adaptedProof: Not applicableverified 2026-08-06 (claude-opus-5)Open item page →

The ordered triple (a,b,c):=((a,b),c)(a,b,c) := ((a,b),c) and the iterated products A×B×C:=(A×B)×CA \times B \times C := (A \times B) \times C

Definition

For sets aa, bb, cc the ordered triple is

(a,b,c):=((a,b),c),(a,b,c) := ((a,b),c),

an ordered pair (The Kuratowski ordered pair (a,b):={{a},{a,b}}(a,b) := \{\{a\},\{a,b\}\}) whose first coordinate is itself an ordered pair, and for sets AA, BB, CC the iterated product is

A×B×C:=(A×B)×C,A \times B \times C := (A \times B) \times C,

a Cartesian product (The Cartesian product A×B:={zP(P(AB)):aA bB 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) \,\}) of the same shape. Both conventions associate to the left, and the elements of A×B×CA \times B \times C are exactly the triples (a,b,c)(a,b,c) with aAa \in A, bBb \in B and cCc \in C.

Applying (a,b)=(c,d)(a,b) = (c,d) if and only if a=ca = c and b=db = d twice gives the characterising property of triples: (a,b,c)=(a,b,c)(a,b,c) = (a',b',c') holds if and only if a=aa = a', b=bb = b' and c=cc = c'.

Remarks

  • The bracketing convention is not a formality. (A×B)×C(A \times B) \times C and A×(B×C)A \times (B \times C) are in general different sets, so a convention has to be fixed and adhered to; the left-associated one is fixed here.

  • No nn-tuples. A general nn-tuple is a function on a natural number, and the natural numbers are not available at this point in the reading order, so only triples and finitely iterated binary products are introduced here. The general construction is the product of an indexed family, once functions and index sets are available.

RemarkRemark: AI-adaptedProof: Not applicableverified 2026-08-06 (claude-opus-5)Open item page →

The axiom ledger for this page: which of the ZFC axioms each construction and each result actually consumes

Remark

Each construction here is licensed by named axioms, and this is the account of which ones.

Two redundancies in the list are worth recording, because they explain why the ledger above is not the shortest possible one. Separation is derivable from Replacement together with a set that has no elements, by a derivation that splits on whether the separating condition holds anywhere in the ambient set; The Axiom Schema of Replacement: for each formula φ\varphi, if φ\varphi defines a class function on AA then its image on AA is a set carries it. Pairing is derivable from Replacement applied to a set with at least two elements, such as the double power set of the empty set. Both are nevertheless assumed here, because each is used constantly and deriving it every time would obscure what a construction actually costs.

The empty set runs the other way. It is often taken as an axiom, and here it is derived instead, from Separation together with the logical fact that the domain of discourse is nonempty; There is exactly one set with no elements, written \varnothing carries the derivation.

5 · Examples, counterexamples and false statements

None yet.

Sources

Standard references

Recommended treatments; not extraction sources.