Metamath Proof Explorer - mmtheorems36

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Statement List for Metamath Proof Explorer - 3501-3600 - Page 36 of 58

Type	Label	Description
Statement
Theorem	r1ord3 3501	Ordering relation for the cumulative hierarchy of sets. Part of Theorem 3.3(i) of [BellMachover] p. 478.
⊢ ((A ∈ On ∧ B ∈ On) → (A ⊆ B → (R1 ‘A) ⊆ (R1 ‘B)))
Theorem	r1val1 3502	The value of the cumulative hierarchy of sets function expressed recursively. Theorem 7Q of [Enderton] p. 202.
⊢ (A ∈ On → (R1 ‘A) = ∪x ∈ A ℘(R1 ‘x))
Theorem	tz9.12lem1 3503	Lemma for tz9.12 3506.
⊢ A ∈ V    &   ⊢ F = {⟨z, w⟩∣w = ∩{v ∈ On∣z ∈ (R1 ‘v)}}    ⇒   ⊢ (F “ A) ⊆ On
Theorem	tz9.12lem2 3504	Lemma for tz9.12 3506.
⊢ A ∈ V    &   ⊢ F = {⟨z, w⟩∣w = ∩{v ∈ On∣z ∈ (R1 ‘v)}}    ⇒   ⊢ suc ∪(F “ A) ∈ On
Theorem	tz9.12lem3 3505	Lemma for tz9.12 3506.
⊢ A ∈ V    &   ⊢ F = {⟨z, w⟩∣w = ∩{v ∈ On∣z ∈ (R1 ‘v)}}    ⇒   ⊢ (∀x ∈ A ∃y ∈ On x ∈ (R1 ‘y) → A ∈ (R1 ‘suc suc ∪(F “ A)))
Theorem	tz9.12 3506	A set is well-founded if all of its elements are well-founded. Proposition 9.12 of [TakeutiZaring] p. 78. The main proof consists of tz9.12lem1 3503 through tz9.12lem3 3505.
⊢ A ∈ V    ⇒   ⊢ (∀x ∈ A ∃y ∈ On x ∈ (R1 ‘y) → ∃y ∈ On A ∈ (R1 ‘y))
Theorem	tz9.13 3507	Every set is well-founded, assuming the Axiom of Regularity. In other words, every set belongs to a layer of the cumulative hierarchy of sets. Proposition 9.13 of [TakeutiZaring] p. 78.
⊢ A ∈ V    ⇒   ⊢ ∃x ∈ On A ∈ (R1 ‘x)
Theorem	tz9.13g 3508	Every set is well-founded, assuming the Axiom of Regularity. Proposition 9.13 of [TakeutiZaring] p. 78. This variant of tz9.13 3507 expresses the class existence requirement as an antecedent.
⊢ (A ∈ B → ∃x ∈ On A ∈ (R1 ‘x))
Theorem	rankwflem 3509	Every set is well-founded, assuming the Axiom of Regularity. Proposition 9.13 of [TakeutiZaring] p. 78. This variant of tz9.13g 3508 is useful in proofs of theorems about the rank function.
⊢ (A ∈ B → ∃x ∈ On A ∈ (R1 ‘suc x))
Theorem	jech9.3 3510	Every set belongs to some value of the cumulative hierarchy of sets function R1, i.e. the indexed union of all values of R1 is the universe. Lemma 9.3 of [Jech] p. 71.
⊢ ∪x ∈ On (R1 ‘x) = V
Theorem	unir1 3511	The cumulative hierarchy of sets covers the universe. Proposition 4.45 (b) to (a) of [Mendelson] p. 281.
⊢ ∪(R1 “ On) = V
Theorem	rankval 3512	Value of the rank function. Definition 9.14 of [TakeutiZaring] p. 79 (proved as a theorem from our definition).
⊢ A ∈ V    ⇒   ⊢ (rank ‘A) = ∩{x ∈ On∣A ∈ (R1 ‘suc x)}
Theorem	rankvalg 3513	Value of the rank function. Definition 9.14 of [TakeutiZaring] p. 79 (proved as a theorem from our definition). This more general variant of rankval 3512 expresses the class existence requirement as an antecedent instead of a hypothesis.
⊢ (A ∈ B → (rank ‘A) = ∩{x ∈ On∣A ∈ (R1 ‘suc x)})
Theorem	rankval2 3514	Value of an alternate definition of the rank function. Definition of [BellMachover] p. 478.
⊢ (A ∈ B → (rank ‘A) = ∩{x ∈ On∣A ⊆ (R1 ‘x)})
Theorem	rankon 3515	The rank of a set is an ordinal number. Proposition 9.15(1) of [TakeutiZaring] p. 79.
⊢ (rank ‘A) ∈ On
Theorem	rankid 3516	Identity law for the rank function.
⊢ A ∈ V    ⇒   ⊢ A ∈ (R1 ‘suc (rank ‘A))
Theorem	rankr1lem 3517	Lemma for rankr1 3518.
⊢ A ∈ V    ⇒   ⊢ (B ∈ On → (¬ A ∈ (R1 ‘B) → B ⊆ (rank ‘A)))
Theorem	rankr1 3518	A relationship between the rank function and the cumulative hierarchy of sets function R1. Proposition 9.15(2) of [TakeutiZaring] p. 79.
⊢ A ∈ V    ⇒   ⊢ (B = (rank ‘A) ↔ (¬ A ∈ (R1 ‘B) ∧ A ∈ (R1 ‘suc B)))
Theorem	rankr1g 3519	A relationship between the rank function and the cumulative hierarchy of sets function R1. Proposition 9.15(2) of [TakeutiZaring] p. 79. This more general variant of rankr1 3518 expresses the class existence requirement as an antecedent.
⊢ (A ∈ C → (B = (rank ‘A) ↔ (¬ A ∈ (R1 ‘B) ∧ A ∈ (R1 ‘suc B))))
Theorem	ssrankr1 3520	A relationship between an ordinal number less than or equal to a rank, and the cumulative hierarchy of sets R1. Proposition 9.15(3) of [TakeutiZaring] p. 79.
⊢ A ∈ V    ⇒   ⊢ (B ∈ On → (B ⊆ (rank ‘A) ↔ ¬ A ∈ (R1 ‘B)))
Theorem	rankr1a 3521	A relationship between rank and R1, clearly equivalent to ssrankr1 3520 and friends through trichotomy, but in Raph's opinion considerably more intuitive. (Contributed by Raph Levien, 29-May-04.)
⊢ A ∈ V    ⇒   ⊢ (B ∈ On → (A ∈ (R1 ‘B) ↔ (rank ‘A) ∈ B))
Theorem	r1val2 3522	The value of the cumulative hierarchy of sets function expressed in terms of rank. Definition 15.19 of [Monk1] p. 113.
⊢ (A ∈ On → (R1 ‘A) = {x∣(rank ‘x) ∈ A})
Theorem	r1val3 3523	The value of the cumulative hierarchy of sets function expressed in terms of rank. Theorem 15.18 of [Monk1] p. 113.
⊢ (A ∈ On → (R1 ‘A) = ∪x ∈ A ℘{y∣(rank ‘y) ∈ x})
Theorem	rankel 3524	The membership relation is inherited by the rank function. Proposition 9.16 of [TakeutiZaring] p. 79.
⊢ B ∈ V    ⇒   ⊢ (A ∈ B → (rank ‘A) ∈ (rank ‘B))
Theorem	rankval3 3525	The value of the rank function expressed recursively: the rank of a set is the smallest ordinal number containing the ranks of all members of the set. Proposition 9.17 of [TakeutiZaring] p. 79.
⊢ A ∈ V    ⇒   ⊢ (rank ‘A) = ∩{x ∈ On∣∀y ∈ A (rank ‘y) ∈ x}
Theorem	bndrank 3526	Any class whose elements have bounded rank is a set. Proposition 9.19 of [TakeutiZaring] p. 80.
⊢ (∃x ∈ On ∀y ∈ A (rank ‘y) ⊆ x → A ∈ V)
Theorem	unbndrank 3527	The elements of a proper class have unbounded rank. Exercise 2 of [TakeutiZaring] p. 80.
⊢ (¬ A ∈ V → ∀x ∈ On ∃y ∈ A x ∈ (rank ‘y))
Theorem	rankpw 3528	The rank of a power set. Part of Exercise 30 of [Enderton] p. 207.
⊢ A ∈ V    ⇒   ⊢ (rank ‘℘A) = suc (rank ‘A)
Theorem	r1pw 3529	A stronger property of R1 than rankpw 3528. The latter merely proves that R1 of the successor is a powerset, but here we prove that if A is in the cumulative hierarchy, then ℘A is in the cumulative hierarchy of the successor. (Contributed by Raph Levien, 29-May-04.)
⊢ (B ∈ On → (A ∈ (R1 ‘B) ↔ ℘A ∈ (R1 ‘suc B)))
Theorem	r1pwcl 3530	The cumulative hierarchy of a limit ordinal is closed under powerset. (Contributed by Raph Levien, 29-May-04.)
⊢ (Lim B → (A ∈ (R1 ‘B) ↔ ℘A ∈ (R1 ‘B)))
Theorem	rankss 3531	The subset relation is inherited by the rank function. Exercise 1 of [TakeutiZaring] p. 80.
⊢ B ∈ V    ⇒   ⊢ (A ⊆ B → (rank ‘A) ⊆ (rank ‘B))
Theorem	ranksn 3532	The rank of a singleton. Theorem 15.17(v) of [Monk1] p. 112.
⊢ A ∈ V    ⇒   ⊢ (rank ‘{A}) = suc (rank ‘A)
Theorem	rankuni 3533	The rank of a union. Part of Theorem 15.17(iv) of [Monk1] p. 112.
⊢ A ∈ V    ⇒   ⊢ (rank ‘∪A) = ∪x ∈ A (rank ‘x)
Theorem	rankuniss 3534	Upper bound of the rank of a union. Part of Exercise 30 of [Enderton] p. 207.
⊢ A ∈ V    ⇒   ⊢ (rank ‘∪A) ⊆ (rank ‘A)
Theorem	rankun 3535	The rank of the union of two sets. Theorem 15.17(iii) of [Monk1] p. 112.
⊢ A ∈ V    &   ⊢ B ∈ V    ⇒   ⊢ (rank ‘(A ∪ B)) = ((rank ‘A) ∪ (rank ‘B))
Theorem	rankpr 3536	The rank of an unordered pair. Part of Exercise 30 of [Enderton] p. 207.
⊢ A ∈ V    &   ⊢ B ∈ V    ⇒   ⊢ (rank ‘{A, B}) = suc ((rank ‘A) ∪ (rank ‘B))
Theorem	r1rankid 3537	Any set is a subset of the hierarchy of its rank.
⊢ (A ∈ B → A ⊆ (R1 ‘(rank ‘A)))
Theorem	rankonid 3538	The rank of an ordinal number is itself. Proposition 9.18 of [TakeutiZaring] p. 79 and its converse.
⊢ (A ∈ On ↔ (rank ‘A) = A)
Theorem	rankr1id 3539	The rank of the hierarchy of an ordinal number is itself.
⊢ (A ∈ On ↔ (rank ‘(R1 ‘A)) = A)
Theorem	ranklon 3540	The rank of a set is the supremum of the successors of the ranks of its members. Exercise 9.1 of [Jech] p. 72. Also a special case of Theorem 7V(b) of [Enderton] p. 204.
⊢ A ∈ V    ⇒   ⊢ (rank ‘A) = ∪x ∈ A suc (rank ‘x)
Theorem	scottex 3541	Scott's trick collects all sets that have a certain property and are of smallest possible rank. This theorem shows that the resulting collection, expressed as in Equation 9.3 of [Jech] p. 72, is a set.
⊢ {x ∈ A∣∀y ∈ A (rank ‘x) ⊆ (rank ‘y)} ∈ V
Theorem	scott0 3542	Scott's trick collects all sets that have a certain property and are of smallest possible rank. This theorem shows that the resulting collection, expressed as in Equation 9.3 of [Jech] p. 72, contains at least one representative with the property, if there is one. In other words, the collection is empty iff no set has the property (i.e. A is empty).
⊢ (A = ∅ ↔ {x ∈ A∣∀y ∈ A (rank ‘x) ⊆ (rank ‘y)} = ∅)
Theorem	scottexs 3543	Theorem scheme version of scottex 3541. The collection of all x of minimum rank such that φ(x) is true, is a set.
⊢ {x∣(φ ∧ ∀y([y / x]φ → (rank ‘x) ⊆ (rank ‘y)))} ∈ V
Theorem	scott0s 3544	Theorem scheme version of scott0 3542. The collection of all x of minimum rank such that φ(x) is true, is not empty iff there is an x such that φ(x) holds.
⊢ (∃xφ ↔ ¬ {x∣(φ ∧ ∀y([y / x]φ → (rank ‘x) ⊆ (rank ‘y)))} = ∅)
Theorem	cplem1 3545	Lemma for the Collection Principle cp 3547.
⊢ C = {y ∈ B∣∀z ∈ B (rank ‘y) ⊆ (rank ‘z)}    &   ⊢ D = ∪x ∈ A C    ⇒   ⊢ ∀x ∈ A (¬ B = ∅ → ¬ (B ∩ D) = ∅)
Theorem	cplem2 3546	Lemma for the Collection Principle cp 3547.
⊢ A ∈ V    ⇒   ⊢ ∃y∀x ∈ A (¬ B = ∅ → ¬ (B ∩ y) = ∅)
Theorem	cp 3547	Collection Principle. This remarkable theorem scheme is in effect a very strong generalization of the Axiom of Replacement. The proof makes use of Scott's trick scottex 3541 that collapses a proper class into a set of minimum rank. The wff φ can be thought of as φ(x, y). Scheme "Collection Principle" of [Jech] p. 72.
⊢ ∃w∀x ∈ z (∃yφ → ∃y ∈ w φ)
Theorem	bnd 3548	A very strong generalization of the Axiom of Replacement (compare zfrep6 2744), derived from the Collection Principle cp 3547. Its strength lies in the rather profound fact that φ(x, y) does not have to be a "function-like" wff, as it does in the standard Axiom of Replacement. This theorem is sometimes called the Boundedness Axiom.
⊢ (∀x ∈ z ∃yφ → ∃w∀x ∈ z ∃y ∈ w φ)
Theorem	bnd2 3549	A variant of the Boundedness Axiom bnd 3548 that picks a subset z out of a possibly proper class B in which a property is true.
⊢ A ∈ V    ⇒   ⊢ (∀x ∈ A ∃y ∈ B φ → ∃z(z ⊆ B ∧ ∀x ∈ A ∃y ∈ z φ))
Theorem	kardex 3550	The collection of all sets equinumerous to a set A and having least possible rank is a set. This is the part of the justification of the definition of kard of [Enderton] p. 222.
⊢ A ∈ V    ⇒   ⊢ {x∣(x ≈ A ∧ ∀y(y ≈ A → (rank ‘x) ⊆ (rank ‘y)))} ∈ V
Theorem	karden 3551	If we allow the Axiom of Regularity, we can avoid the Axiom of Choice by defining the cardinal number of a set as the set of all sets equinumerous to it and having least possible rank. This theorem proves the equinumerosity relationship for this definition (compare carden 3638). The hypotheses correspond to the definition of kard of [Enderton] p. 222 (which we don't define separately since currently we do not use it elsewhere). This theorem along with kardex 3550 justify the definition of kard.
⊢ A ∈ V    &   ⊢ B ∈ V    &   ⊢ C = {x∣(x ≈ A ∧ ∀y(y ≈ A → (rank ‘x) ⊆ (rank ‘y)))}    &   ⊢ D = {x∣(x ≈ B ∧ ∀y(y ≈ B → (rank ‘x) ⊆ (rank ‘y)))}    ⇒   ⊢ (C = D ↔ A ≈ B)
Theorem	aceq1 3552	Equivalence of two versions of the Axiom of Choice ax-ac 1080. The proof uses neither AC nor the Axiom of Regularity. The right-hand side expresses our AC with the fewest number of different variables.
⊢ (∃y∀z ∈ x ∀w ∈ z ∃!v ∈ z ∃u ∈ y (z ∈ u ∧ v ∈ u) ↔ ∃y∀z∀w((z ∈ w ∧ w ∈ x) → ∃x∀z(∃x((z ∈ w ∧ w ∈ x) ∧ (z ∈ x ∧ x ∈ y)) ↔ z = x)))
Theorem	aceq0 3553	Equivalence of two versions of the Axiom of Choice. The proof uses neither AC nor the Axiom of Regularity. The right-hand side is our original ax-ac 1080.
⊢ (∃y∀z ∈ x ∀w ∈ z ∃!v ∈ z ∃u ∈ y (z ∈ u ∧ v ∈ u) ↔ ∃y∀z∀w((z ∈ w ∧ w ∈ x) → ∃v∀u(∃t((u ∈ w ∧ w ∈ t) ∧ (u ∈ t ∧ t ∈ y)) ↔ u = v)))
Theorem	aceq2 3554	Equivalence of two versions of the Axiom of Choice. The proof uses neither AC nor the Axiom of Regularity.
⊢ (∃y∀z ∈ x ∀w ∈ z ∃!v ∈ z ∃u ∈ y (z ∈ u ∧ v ∈ u) ↔ ∃y∀z ∈ x (¬ z = ∅ → ∃!w ∈ z ∃v ∈ y (z ∈ v ∧ w ∈ v)))
Theorem	aceq3lem 3555	Lemma for aceq3 3556.
⊢ F = {⟨w, v⟩∣(w ∈ dom y ∧ v = (f ‘{u∣wyu}))}    ⇒   ⊢ (∀x∃f∀z ∈ x (¬ z = ∅ → (f ‘z) ∈ z) → ∃f(f ⊆ y ∧ f Fn dom y))
Theorem	aceq3 3556	Equivalence of two versions of the Axiom of Choice. The left-hand side is similar to the Axiom of Choice (first form) of [Enderton] p. 49. The right-hand side is the Axiom of Choice of [TakeutiZaring] p. 83. The proof does not depend on AC.
⊢ (∀x∃f(f ⊆ x ∧ f Fn dom x) ↔ ∀x∃f∀z ∈ x (¬ z = ∅ → (f ‘z) ∈ z))
Theorem	aceq4 3557	Equivalence of two versions of the Axiom of Choice. The left-hand side is similar to the Axiom of Choice (first form) of [Enderton] p. 49. The right-hand side is Axiom AC of [BellMachover] p. 488. The proof does not depend on AC.
⊢ (∀x∃f(f ⊆ x ∧ f Fn dom x) ↔ ∀x∃f(f Fn x ∧ ∀z ∈ x (¬ z = ∅ → (f ‘z) ∈ z)))
Theorem	aceq5lem1 3558	Lemma for aceq5 3563.
⊢ (∃!v v ∈ (({w} × w) ∩ y) ↔ ∃!g(g ∈ w ∧ ⟨w, g⟩ ∈ y))
Theorem	aceq5lem2 3559	Lemma for aceq5 3563.
⊢ A = {u∣(¬ u = ∅ ∧ ∃t ∈ h u = ({t} × t))}    ⇒   ⊢ (⟨w, g⟩ ∈ ∪A ↔ (w ∈ h ∧ g ∈ w))
Theorem	aceq5lem3 3560	Lemma for aceq5 3563.
⊢ A = {u∣(¬ u = ∅ ∧ ∃t ∈ h u = ({t} × t))}    ⇒   ⊢ (({w} × w) ∈ A ↔ (¬ w = ∅ ∧ w ∈ h))
Theorem	aceq5lem4 3561	Lemma for aceq5 3563.
⊢ A = {u∣(¬ u = ∅ ∧ ∃t ∈ h u = ({t} × t))}    &   ⊢ B = (∪A ∩ y)    &   ⊢ (φ ↔ ∀x((∀z ∈ x ¬ z = ∅ ∧ ∀z ∈ x ∀w ∈ x (¬ z = w → (z ∩ w) = ∅)) → ∃y∀z ∈ x ∃!v v ∈ (z ∩ y)))    ⇒   ⊢ (φ → ∃y∀z ∈ A ∃!v v ∈ (z ∩ y))
Theorem	aceq5lem5 3562	Lemma for aceq5 3563.
⊢ A = {u∣(¬ u = ∅ ∧ ∃t ∈ h u = ({t} × t))}    &   ⊢ B = (∪A ∩ y)    &   ⊢ (φ ↔ ∀x((∀z ∈ x ¬ z = ∅ ∧ ∀z ∈ x ∀w ∈ x (¬ z = w → (z ∩ w) = ∅)) → ∃y∀z ∈ x ∃!v v ∈ (z ∩ y)))    ⇒   ⊢ (φ → ∃f∀w ∈ h (¬ w = ∅ → (f ‘w) ∈ w))
Theorem	aceq5 3563	Equivalence of two versions of the Axiom of Choice. The left-hand side is similar to the Axiom of Choice (first form) of [Enderton] p. 49. The right-hand side is Theorem 6M(4) of [Enderton] p. 151 and asserts that given a family of mutually disjoint nonempty sets, a set exists containing exactly one member from each set in the family. The proof does not depend on AC.
⊢ (∀x∃f(f ⊆ x ∧ f Fn dom x) ↔ ∀x((∀z ∈ x ¬ z = ∅ ∧ ∀z ∈ x ∀w ∈ x (¬ z = w → (z ∩ w) = ∅)) → ∃y∀z ∈ x ∃!v v ∈ (z ∩ y)))
Theorem	aceq6a 3564	Our Axiom of Choice (in the form of ac3 3568) implies the Axiom of Choice (first form) of [Enderton] p. 49. The proof uses neither AC nor the Axiom of Regularity. See aceq6b 3565 for the converse (which does use the Axiom of Regularity).
⊢ (∀x∃y∀z ∈ x (¬ z = ∅ → ∃!w ∈ z ∃v ∈ y (z ∈ v ∧ w ∈ v)) → ∀x∃f(f ⊆ x ∧ f Fn dom x))
Theorem	aceq6b 3565	Axiom of Choice (first form) of [Enderton] p. 49 implies of our Axiom of Choice (in the form of ac3 3568). The proof does not make use of AC. Note that the Axiom of Regularity is used by the proof. Specifically, eirrv 3449 and preleq 3454 that are referenced in the proof each make use of Regularity for their derivations. (The reverse derivation can be done without using Regularity; see aceq6a 3564.)
⊢ (∀x∃f(f ⊆ x ∧ f Fn dom x) → ∀x∃y∀z ∈ x (¬ z = ∅ → ∃!w ∈ z ∃v ∈ y (z ∈ v ∧ w ∈ v)))
Theorem	aceq7 3566	Equivalence of the Axiom of Choice (first form) of [Enderton] p. 49 and our Axiom of Choice (in the form of ac2 3567). The proof does not depend AC on but does depend on the Axiom of Regularity.
⊢ (∀x∃f(f ⊆ x ∧ f Fn dom x) ↔ ∀x∃y∀z ∈ x ∀w ∈ z ∃!v ∈ z ∃u ∈ y (z ∈ u ∧ v ∈ u))
Theorem	ac2 3567	Axiom of Choice using abbreviations. This cute and very short version does not make use of any defined objects such as the empty set or a function value. However, it is hard to explain intuitively. If you want to figure it out, the rewritten equivalent ac3 3568 is easier to understand. Note: aceq0 3553 shows the logical equivalence to ax-ac 1080.
⊢ ∃y∀z ∈ x ∀w ∈ z ∃!v ∈ z ∃u ∈ y (z ∈ u ∧ v ∈ u)
Theorem	ac3 3568	Axiom of Choice using abbreviations. The logical equivalence to ax-ac 1080 can be established by chaining aceq0 3553 and aceq2 3554. A standard textbook version of AC is derived from this one in aceq6a 3564, and this version of AC is derived from the textbook version in aceq6b 3565. The following sketch will help you understand this version of the axiom. Given any set x, the axiom says that there exists a y that is a collection of unordered pairs, one pair for each non-empty member of x. One entry in the pair is the member of x, and the other entry is some arbitrary member of that member of x. Using the Axiom of Regularity, we can show that y is really a set of ordered pairs, very similar to the ordered pair construction opthreg 3455. The key theorem for this (used in the proof of aceq6b 3565) is preleq 3454. With this modified definition of ordered pair, it can be seen that y is actually a choice function on the members of x. For example, suppose x = {{1, 2}, {1, 3}, {2, 3}}. Take y = {{{1, 2}, 1}, {{1, 3}, 1}, {{2, 3}, 2}}. For the member (of x) z = {1, 2}, the only assignment to w and v that satisfies the axiom is w = 1 and v = {{1, 2}, 1}, so there is exactly one w as required. We verify the other two members of x similarly. Thus y satisfies the axiom. Using our modified ordered pair definition, it is easy to see that y is the choice function {⟨{1, 2}, 1⟩, ⟨{1, 3}, 1⟩, ⟨{2, 3}, 2⟩}. Of course other choices for y will also satisfy the axiom, for example y = {{{1, 2}, 2}, {{1, 3}, 1}, {{2, 3}, 3}}. What AC tells us is that there exists at least one such y, but it doesn't tell us which one.
⊢ ∃y∀z ∈ x (¬ z = ∅ → ∃!w ∈ z ∃v ∈ y (z ∈ v ∧ w ∈ v))
Theorem	ac7 3569	An Axiom of Choice equivalent similar to the Axiom of Choice (first form) of [Enderton] p. 49.
⊢ ∃f(f ⊆ x ∧ f Fn dom x)
Theorem	ac7g 3570	An Axiom of Choice equivalent similar to the Axiom of Choice (first form) of [Enderton] p. 49.
⊢ (R ∈ A → ∃f(f ⊆ R ∧ f Fn dom R))
Theorem	ac4 3571	Equivalent of Axiom of Choice. We do not insist that f be a function. However, theorem ac5 3573, derived from this one, shows that this form of the axiom does imply that at least one such set f whose existence we assert is in fact a function. Axiom of Choice of [TakeutiZaring] p. 83.
⊢ ∃f∀z ∈ x (¬ z = ∅ → (f ‘z) ∈ z)
Theorem	ac4c 3572	Equivalent of Axiom of Choice (class version)
⊢ A ∈ V    ⇒   ⊢ ∃f∀x ∈ A (¬ x = ∅ → (f ‘x) ∈ x)
Theorem	ac5 3573	An Axiom of Choice equivalent. Axiom AC of [BellMachover] p. 488.
⊢ A ∈ V    ⇒   ⊢ ∃f(f Fn A ∧ ∀x ∈ A (¬ x = ∅ → (f ‘x) ∈ x))
Theorem	ac5b 3574	Equivalent of Axiom of Choice.
⊢ A ∈ V    ⇒   ⊢ (∀x ∈ A ¬ x = ∅ → ∃f(f:A–→∪A ∧ ∀x ∈ A (f ‘x) ∈ x))
Theorem	ac6lem 3575	Lemma for equivalent of Axiom of Choice.
⊢ A ∈ V    &   ⊢ B ∈ V    &   ⊢ C = {y ∈ B∣φ}    &   ⊢ H = {⟨x, z⟩∣(x ∈ A ∧ z = C)}    ⇒   ⊢ (∀x ∈ A ∃y ∈ B φ → ∃f(f:A–→B ∧ ∀x ∈ A (f ‘x) ∈ C))
Theorem	ac6 3576	Equivalent of Axiom of Choice. This is useful for proving that there exists, for example, a sequence mapping natural numbers to members of a large set B, where φ depends on x (the natural number) and y (to specify a member of B). A stronger version of this theorem, ac6s 3577, allows B to be a proper class.
⊢ A ∈ V    &   ⊢ B ∈ V    &   ⊢ (y = (f ‘x) → (φ ↔ ψ))    ⇒   ⊢ (∀x ∈ A ∃y ∈ B φ → ∃f(f:A–→B ∧ ∀x ∈ A ψ))
Theorem	ac6s 3577	Equivalent of Axiom of Choice. Using the Boundedness Axiom bnd2 3549, we derive this strong version of ac6 3576 that doesn't require B to be a set.
⊢ A ∈ V    &   ⊢ (y = (f ‘x) → (φ ↔ ψ))    ⇒   ⊢ (∀x ∈ A ∃y ∈ B φ → ∃f(f:A–→B ∧ ∀x ∈ A ψ))
Theorem	ac6s2 3578	Generalization of the Axiom of Choice to classes. Theorem 10.46 of [TakeutiZaring] p. 97.
⊢ A ∈ V    &   ⊢ (y = (f ‘x) → (φ ↔ ψ))    ⇒   ⊢ (∀x ∈ A ∃yφ → ∃f∀x ∈ A ψ)
Theorem	ac8 3579	An Axiom of Choice equivalent. Given a family x of mutually disjoint nonempty sets, there exists a set y containing exactly one member from each set in the family. Theorem 6M(4) of [Enderton] p. 151.
⊢ ((∀z ∈ x ¬ z = ∅ ∧ ∀z ∈ x ∀w ∈ x (¬ z = w → (z ∩ w) = ∅)) → ∃y∀z ∈ x ∃!v v ∈ (z ∩ y))
Theorem	kmlem1 3580	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, 1 => 2.
⊢ (∀x((∀z ∈ x ¬ z = ∅ ∧ ∀z ∈ x ∀w ∈ x φ) → ∃y∀z ∈ x ψ) → ∀x(∀z ∈ x ∀w ∈ x φ → ∃y∀z ∈ x (¬ z = ∅ → ψ)))
Theorem	kmlem2 3581	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4.
⊢ (∃y∀z ∈ x (φ → ∃!v v ∈ (z ∩ y)) ↔ ∃y(¬ y ∈ x ∧ ∀z ∈ x (φ → ∃!v v ∈ (z ∩ y))))
Theorem	kmlem3 3582	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4. The right-hand side is part of the hypothesis of 4.
⊢ (¬ (z ∖ ∪(x ∖ {z})) = ∅ ↔ ∃v ∈ z ∀w ∈ x (¬ z = w → ¬ v ∈ (z ∩ w)))
Theorem	kmlem4 3583	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4.
⊢ ((w ∈ x ∧ ¬ z = w) → ((z ∖ ∪(x ∖ {z})) ∩ w) = ∅)
Theorem	kmlem5 3584	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4.
⊢ ((w ∈ x ∧ ¬ z = w) → ((z ∖ ∪(x ∖ {z})) ∩ (w ∖ ∪(x ∖ {w}))) = ∅)
Theorem	kmlem6 3585	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 4 => 1.
⊢ ((∀z ∈ x ¬ z = ∅ ∧ ∀z ∈ x ∀w ∈ x (φ → A = ∅)) → ∀z ∈ x ∃v ∈ z ∀w ∈ x (φ → ¬ v ∈ A))
Theorem	kmlem7 3586	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 4 => 1.
⊢ ((∀z ∈ x ¬ z = ∅ ∧ ∀z ∈ x ∀w ∈ x (¬ z = w → (z ∩ w) = ∅)) → ¬ ∃z ∈ x ∀v ∈ z ∃w ∈ x (¬ z = w ∧ v ∈ (z ∩ w)))
Theorem	kmlem8 3587	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4.
⊢ A = {u∣∃t ∈ x u = (t ∖ ∪(x ∖ {t}))}    ⇒   ⊢ ∀z ∈ A ∀w ∈ A (¬ z = w → (z ∩ w) = ∅)
Theorem	kmlem9 3588	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4.
⊢ A = {u∣∃t ∈ x u = (t ∖ ∪(x ∖ {t}))}    ⇒   ⊢ (∀h(∀z ∈ h ∀w ∈ h (¬ z = w → (z ∩ w) = ∅) → ∃y∀z ∈ h φ) → ∃y∀z ∈ A φ)
Theorem	kmlem10 3589	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4.
⊢ A = {u∣∃t ∈ x u = (t ∖ ∪(x ∖ {t}))}    ⇒   ⊢ (z ∈ x → (z ∩ ∪A) = (z ∖ ∪(x ∖ {z})))
Theorem	kmlem11 3590	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4.
⊢ A = {u∣∃t ∈ x u = (t ∖ ∪(x ∖ {t}))}    ⇒   ⊢ (∀z ∈ x ¬ (z ∖ ∪(x ∖ {z})) = ∅ → (∀z ∈ A (¬ z = ∅ → ∃!v v ∈ (z ∩ y)) → ∀z ∈ x (¬ z = ∅ → ∃!v v ∈ (z ∩ (y ∩ ∪A)))))
Theorem	kmlem12 3591	Lemma for 5-quantifier AC of Kurt Maes, Th. 4 1 <=> 4.
⊢ A = {u∣∃t ∈ x u = (t ∖ ∪(x ∖ {t}))}    ⇒   ⊢ (∀x((∀z ∈ x ¬ z = ∅ ∧ ∀z ∈ x ∀w ∈ x (¬ z = w → (z ∩ w) = ∅)) → ∃y∀z ∈ x ∃!v v ∈ (z ∩ y)) ↔ ∀x(¬ ∃z ∈ x ∀v ∈ z ∃w ∈ x (¬ z = w ∧ v ∈ (z ∩ w)) → ∃y∀z ∈ x (¬ z = ∅ → ∃!v v ∈ (z ∩ y))))
Theorem	kmlem13 3592	Lemma for 5-quantifier AC of Kurt Maes, Th. 4 1 <=> 4.
⊢ ((¬ ∃z ∈ x ∀v ∈ z φ → ∃y∀z ∈ x (¬ z = ∅ → ∃!v v ∈ (z ∩ y))) ↔ (∃z ∈ x ∀v ∈ z φ ∨ ∃y(¬ y ∈ x ∧ ∀z ∈ x ∃!v v ∈ (z ∩ y))))
Theorem	kmlem14 3593	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 5 <=> 4.
⊢ (φ ↔ (z ∈ y → ((v ∈ x ∧ ¬ y = v) ∧ z ∈ v)))    &   ⊢ (ψ ↔ (z ∈ x → ((v ∈ z ∧ v ∈ y) ∧ ((u ∈ z ∧ u ∈ y) → u = v))))    &   ⊢ (χ ↔ ∀z ∈ x ∃!v v ∈ (z ∩ y))    ⇒   ⊢ (∃z ∈ x ∀v ∈ z ∃w ∈ x (¬ z = w ∧ v ∈ (z ∩ w)) ↔ ∃y∀z∃v∀u(y ∈ x ∧ φ))
Theorem	kmlem15 3594	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 5 <=> 4.
⊢ (φ ↔ (z ∈ y → ((v ∈ x ∧ ¬ y = v) ∧ z ∈ v)))    &   ⊢ (ψ ↔ (z ∈ x → ((v ∈ z ∧ v ∈ y) ∧ ((u ∈ z ∧ u ∈ y) → u = v))))    &   ⊢ (χ ↔ ∀z ∈ x ∃!v v ∈ (z ∩ y))    ⇒   ⊢ ((¬ y ∈ x ∧ χ) ↔ ∀z∃v∀u(¬ y ∈ x ∧ ψ))
Theorem	kmlem16 3595	Lemma for 5-quantifier AC of Kurt Maes, Th. 4 5 <=> 4.
⊢ (φ ↔ (z ∈ y → ((v ∈ x ∧ ¬ y = v) ∧ z ∈ v)))    &   ⊢ (ψ ↔ (z ∈ x → ((v ∈ z ∧ v ∈ y) ∧ ((u ∈ z ∧ u ∈ y) → u = v))))&n