P. 101 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 10001-10100   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	alephsucdom 10001	A set dominated by an aleph is strictly dominated by its successor aleph and vice-versa. (Contributed by NM, 3-Nov-2003.) (Revised by Mario Carneiro, 2-Feb-2013.)
⊢ (𝐵 ∈ On → (𝐴 ≼ (ℵ‘𝐵) ↔ 𝐴 ≺ (ℵ‘suc 𝐵)))
Theorem	alephsuc2 10002*	An alternate representation of a successor aleph. The aleph function is the function obtained from the hartogs 9459 function by transfinite recursion, starting from ω. Using this theorem we could define the aleph function with {𝑧 ∈ On ∣ 𝑧 ≼ 𝑥} in place of ∩ {𝑧 ∈ On ∣ 𝑥 ≺ 𝑧} in df-aleph 9864. (Contributed by NM, 3-Nov-2003.) (Revised by Mario Carneiro, 2-Feb-2013.)
⊢ (𝐴 ∈ On → (ℵ‘suc 𝐴) = {𝑥 ∈ On ∣ 𝑥 ≼ (ℵ‘𝐴)})
Theorem	alephdom 10003	Relationship between inclusion of ordinal numbers and dominance of infinite initial ordinals. (Contributed by Jeff Hankins, 23-Oct-2009.)
⊢ ((𝐴 ∈ On ∧ 𝐵 ∈ On) → (𝐴 ⊆ 𝐵 ↔ (ℵ‘𝐴) ≼ (ℵ‘𝐵)))
Theorem	alephgeom 10004	Every aleph is greater than or equal to the set of natural numbers. (Contributed by NM, 11-Nov-2003.)
⊢ (𝐴 ∈ On ↔ ω ⊆ (ℵ‘𝐴))
Theorem	alephislim 10005	Every aleph is a limit ordinal. (Contributed by NM, 11-Nov-2003.)
⊢ (𝐴 ∈ On ↔ Lim (ℵ‘𝐴))
Theorem	aleph11 10006	The aleph function is one-to-one. (Contributed by NM, 3-Aug-2004.)
⊢ ((𝐴 ∈ On ∧ 𝐵 ∈ On) → ((ℵ‘𝐴) = (ℵ‘𝐵) ↔ 𝐴 = 𝐵))
Theorem	alephf1 10007	The aleph function is a one-to-one mapping from the ordinals to the infinite cardinals. See also alephf1ALT 10025. (Contributed by Mario Carneiro, 2-Feb-2013.)
⊢ ℵ:On–1-1→On
Theorem	alephsdom 10008	If an ordinal is smaller than an initial ordinal, it is strictly dominated by it. (Contributed by Jeff Hankins, 24-Oct-2009.) (Proof shortened by Mario Carneiro, 20-Sep-2014.)
⊢ ((𝐴 ∈ On ∧ 𝐵 ∈ On) → (𝐴 ∈ (ℵ‘𝐵) ↔ 𝐴 ≺ (ℵ‘𝐵)))
Theorem	alephdom2 10009	A dominated initial ordinal is included. (Contributed by Jeff Hankins, 24-Oct-2009.)
⊢ ((𝐴 ∈ On ∧ 𝐵 ∈ On) → ((ℵ‘𝐴) ⊆ 𝐵 ↔ (ℵ‘𝐴) ≼ 𝐵))
Theorem	alephle 10010	The argument of the aleph function is less than or equal to its value. Exercise 2 of [TakeutiZaring] p. 91. (Later, in alephfp2 10031, we will that equality can sometimes hold.) (Contributed by NM, 9-Nov-2003.) (Proof shortened by Mario Carneiro, 22-Feb-2013.)
⊢ (𝐴 ∈ On → 𝐴 ⊆ (ℵ‘𝐴))
Theorem	cardaleph 10011*	Given any transfinite cardinal number 𝐴, there is exactly one aleph that is equal to it. Here we compute that aleph explicitly. (Contributed by NM, 9-Nov-2003.) (Revised by Mario Carneiro, 2-Feb-2013.)
⊢ ((ω ⊆ 𝐴 ∧ (card‘𝐴) = 𝐴) → 𝐴 = (ℵ‘∩ {𝑥 ∈ On ∣ 𝐴 ⊆ (ℵ‘𝑥)}))
Theorem	cardalephex 10012*	Every transfinite cardinal is an aleph and vice-versa. Theorem 8A(b) of [Enderton] p. 213 and its converse. (Contributed by NM, 5-Nov-2003.)
⊢ (ω ⊆ 𝐴 → ((card‘𝐴) = 𝐴 ↔ ∃𝑥 ∈ On 𝐴 = (ℵ‘𝑥)))
Theorem	infenaleph 10013*	An infinite numerable set is equinumerous to an infinite initial ordinal. (Contributed by Jeff Hankins, 23-Oct-2009.) (Revised by Mario Carneiro, 29-Apr-2015.)
⊢ ((𝐴 ∈ dom card ∧ ω ≼ 𝐴) → ∃𝑥 ∈ ran ℵ𝑥 ≈ 𝐴)
Theorem	isinfcard 10014	Two ways to express the property of being a transfinite cardinal. (Contributed by NM, 9-Nov-2003.)
⊢ ((ω ⊆ 𝐴 ∧ (card‘𝐴) = 𝐴) ↔ 𝐴 ∈ ran ℵ)
Theorem	iscard3 10015	Two ways to express the property of being a cardinal number. (Contributed by NM, 9-Nov-2003.)
⊢ ((card‘𝐴) = 𝐴 ↔ 𝐴 ∈ (ω ∪ ran ℵ))
Theorem	cardnum 10016	Two ways to express the class of all cardinal numbers, which consists of the finite ordinals in ω plus the transfinite alephs. (Contributed by NM, 10-Sep-2004.)
⊢ {𝑥 ∣ (card‘𝑥) = 𝑥} = (ω ∪ ran ℵ)
Theorem	alephinit 10017*	An infinite initial ordinal is characterized by the property of being initial - that is, it is a subset of any dominating ordinal. (Contributed by Jeff Hankins, 29-Oct-2009.) (Proof shortened by Mario Carneiro, 20-Sep-2014.)
⊢ ((𝐴 ∈ On ∧ ω ⊆ 𝐴) → (𝐴 ∈ ran ℵ ↔ ∀𝑥 ∈ On (𝐴 ≼ 𝑥 → 𝐴 ⊆ 𝑥)))
Theorem	carduniima 10018	The union of the image of a mapping to cardinals is a cardinal. Proposition 11.16 of [TakeutiZaring] p. 104. (Contributed by NM, 4-Nov-2004.)
⊢ (𝐴 ∈ 𝐵 → (𝐹:𝐴⟶(ω ∪ ran ℵ) → ∪ (𝐹 “ 𝐴) ∈ (ω ∪ ran ℵ)))
Theorem	cardinfima 10019*	If a mapping to cardinals has an infinite value, then the union of its image is an infinite cardinal. Corollary 11.17 of [TakeutiZaring] p. 104. (Contributed by NM, 4-Nov-2004.)
⊢ (𝐴 ∈ 𝐵 → ((𝐹:𝐴⟶(ω ∪ ran ℵ) ∧ ∃𝑥 ∈ 𝐴 (𝐹‘𝑥) ∈ ran ℵ) → ∪ (𝐹 “ 𝐴) ∈ ran ℵ))
Theorem	alephiso 10020	Aleph is an order isomorphism of the class of ordinal numbers onto the class of infinite cardinals. Definition 10.27 of [TakeutiZaring] p. 90. (Contributed by NM, 3-Aug-2004.)
⊢ ℵ Isom E , E (On, {𝑥 ∣ (ω ⊆ 𝑥 ∧ (card‘𝑥) = 𝑥)})
Theorem	alephprc 10021	The class of all transfinite cardinal numbers (the range of the aleph function) is a proper class. Proposition 10.26 of [TakeutiZaring] p. 90. (Contributed by NM, 11-Nov-2003.)
⊢ ¬ ran ℵ ∈ V
Theorem	alephsson 10022	The class of transfinite cardinals (the range of the aleph function) is a subclass of the class of ordinal numbers. (Contributed by NM, 11-Nov-2003.)
⊢ ran ℵ ⊆ On
Theorem	unialeph 10023	The union of the class of transfinite cardinals (the range of the aleph function) is the class of ordinal numbers. (Contributed by NM, 11-Nov-2003.)
⊢ ∪ ran ℵ = On
Theorem	alephsmo 10024	The aleph function is strictly monotone. (Contributed by Mario Carneiro, 15-Mar-2013.)
⊢ Smo ℵ
Theorem	alephf1ALT 10025	Alternate proof of alephf1 10007. (Contributed by Mario Carneiro, 15-Mar-2013.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ ℵ:On–1-1→On
Theorem	alephfplem1 10026	Lemma for alephfp 10030. (Contributed by NM, 6-Nov-2004.)
⊢ 𝐻 = (rec(ℵ, ω) ↾ ω)    ⇒   ⊢ (𝐻‘∅) ∈ ran ℵ
Theorem	alephfplem2 10027*	Lemma for alephfp 10030. (Contributed by NM, 6-Nov-2004.)
⊢ 𝐻 = (rec(ℵ, ω) ↾ ω)    ⇒   ⊢ (𝑤 ∈ ω → (𝐻‘suc 𝑤) = (ℵ‘(𝐻‘𝑤)))
Theorem	alephfplem3 10028*	Lemma for alephfp 10030. (Contributed by NM, 6-Nov-2004.)
⊢ 𝐻 = (rec(ℵ, ω) ↾ ω)    ⇒   ⊢ (𝑣 ∈ ω → (𝐻‘𝑣) ∈ ran ℵ)
Theorem	alephfplem4 10029	Lemma for alephfp 10030. (Contributed by NM, 5-Nov-2004.)
⊢ 𝐻 = (rec(ℵ, ω) ↾ ω)    ⇒   ⊢ ∪ (𝐻 “ ω) ∈ ran ℵ
Theorem	alephfp 10030	The aleph function has a fixed point. Similar to Proposition 11.18 of [TakeutiZaring] p. 104, except that we construct an actual example of a fixed point rather than just showing its existence. See alephfp2 10031 for an abbreviated version just showing existence. (Contributed by NM, 6-Nov-2004.) (Proof shortened by Mario Carneiro, 15-May-2015.)
⊢ 𝐻 = (rec(ℵ, ω) ↾ ω)    ⇒   ⊢ (ℵ‘∪ (𝐻 “ ω)) = ∪ (𝐻 “ ω)
Theorem	alephfp2 10031	The aleph function has at least one fixed point. Proposition 11.18 of [TakeutiZaring] p. 104. See alephfp 10030 for an actual example of a fixed point. Compare the inequality alephle 10010 that holds in general. Note that if 𝑥 is a fixed point, then ℵ‘ℵ‘ℵ‘... ℵ‘𝑥 = 𝑥. (Contributed by NM, 6-Nov-2004.) (Revised by Mario Carneiro, 15-May-2015.)
⊢ ∃𝑥 ∈ On (ℵ‘𝑥) = 𝑥
Theorem	alephval3 10032*	An alternate way to express the value of the aleph function: it is the least infinite cardinal different from all values at smaller arguments. Definition of aleph in [Enderton] p. 212 and definition of aleph in [BellMachover] p. 490 . (Contributed by NM, 16-Nov-2003.)
⊢ (𝐴 ∈ On → (ℵ‘𝐴) = ∩ {𝑥 ∣ ((card‘𝑥) = 𝑥 ∧ ω ⊆ 𝑥 ∧ ∀𝑦 ∈ 𝐴 ¬ 𝑥 = (ℵ‘𝑦))})
Theorem	alephsucpw2 10033	The power set of an aleph is not strictly dominated by the successor aleph. (The Generalized Continuum Hypothesis says they are equinumerous, see gch3 10599 or gchaleph2 10595.) The transposed form alephsucpw 10493 cannot be proven without the AC, and is in fact equivalent to it. (Contributed by Mario Carneiro, 2-Feb-2013.)
⊢ ¬ 𝒫 (ℵ‘𝐴) ≺ (ℵ‘suc 𝐴)
Theorem	mappwen 10034	Power rule for cardinal arithmetic. Theorem 11.21 of [TakeutiZaring] p. 106. (Contributed by Mario Carneiro, 9-Mar-2013.) (Revised by Mario Carneiro, 27-Apr-2015.)
⊢ (((𝐵 ∈ dom card ∧ ω ≼ 𝐵) ∧ (2o ≼ 𝐴 ∧ 𝐴 ≼ 𝒫 𝐵)) → (𝐴 ↑m 𝐵) ≈ 𝒫 𝐵)
Theorem	finnisoeu 10035*	A finite totally ordered set has a unique order isomorphism to a finite ordinal. (Contributed by Stefan O'Rear, 16-Nov-2014.) (Proof shortened by Mario Carneiro, 26-Jun-2015.)
⊢ ((𝑅 Or 𝐴 ∧ 𝐴 ∈ Fin) → ∃!𝑓 𝑓 Isom E , 𝑅 ((card‘𝐴), 𝐴))
Theorem	iunfictbso 10036	Countability of a countable union of finite sets with a strict (not globally well) order fulfilling the choice role. (Contributed by Stefan O'Rear, 16-Nov-2014.)
⊢ ((𝐴 ≼ ω ∧ 𝐴 ⊆ Fin ∧ 𝐵 Or ∪ 𝐴) → ∪ 𝐴 ≼ ω)
2.6.13  Axiom of Choice equivalents
Syntax	wac 10037	Wff for an abbreviation of the axiom of choice.
wff CHOICE
Definition	df-ac 10038*	The expression CHOICE will be used as a readable shorthand for any form of the axiom of choice; all concrete forms are long, cryptic, have dummy variables, or all three, making it useful to have a short name. Similar to the Axiom of Choice (first form) of [Enderton] p. 49. There is a slight problem with taking the exact form of ax-ac 10381 as our definition, because the equivalence to more standard forms (dfac2 10054) requires the Axiom of Regularity, which we often try to avoid. Thus, we take the first of the "textbook forms" as the definition and derive the form of ax-ac 10381 itself as dfac0 10056. (Contributed by Mario Carneiro, 22-Feb-2015.)
⊢ (CHOICE ↔ ∀𝑥∃𝑓(𝑓 ⊆ 𝑥 ∧ 𝑓 Fn dom 𝑥))
Theorem	aceq1 10039*	Equivalence of two versions of the Axiom of Choice ax-ac 10381. The proof uses neither AC nor the Axiom of Regularity. The right-hand side expresses our AC with the fewest number of different variables. (Contributed by NM, 5-Apr-2004.)
⊢ (∃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 10040*	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 10381. (Contributed by NM, 5-Apr-2004.)
⊢ (∃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 10041*	Equivalence of two versions of the Axiom of Choice. The proof uses neither AC nor the Axiom of Regularity. (Contributed by NM, 5-Apr-2004.)
⊢ (∃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 10042*	Lemma for dfac3 10043. (Contributed by NM, 2-Apr-2004.) (Revised by Mario Carneiro, 26-Jun-2015.)
⊢ 𝐹 = (𝑤 ∈ dom 𝑦 ↦ (𝑓‘{𝑢 ∣ 𝑤𝑦𝑢}))    ⇒   ⊢ (∀𝑥∃𝑓∀𝑧 ∈ 𝑥 (𝑧 ≠ ∅ → (𝑓‘𝑧) ∈ 𝑧) → ∃𝑓(𝑓 ⊆ 𝑦 ∧ 𝑓 Fn dom 𝑦))
Theorem	dfac3 10043*	Equivalence of two versions of the Axiom of Choice. The left-hand side is defined as 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. (Contributed by NM, 24-Mar-2004.) (Revised by Stefan O'Rear, 22-Feb-2015.)
⊢ (CHOICE ↔ ∀𝑥∃𝑓∀𝑧 ∈ 𝑥 (𝑧 ≠ ∅ → (𝑓‘𝑧) ∈ 𝑧))
Theorem	dfac4 10044*	Equivalence of two versions of the Axiom of Choice. The right-hand side is Axiom AC of [BellMachover] p. 488. The proof does not depend on AC. (Contributed by NM, 24-Mar-2004.) (Revised by Mario Carneiro, 26-Jun-2015.)
⊢ (CHOICE ↔ ∀𝑥∃𝑓(𝑓 Fn 𝑥 ∧ ∀𝑧 ∈ 𝑥 (𝑧 ≠ ∅ → (𝑓‘𝑧) ∈ 𝑧)))
Theorem	dfac5lem1 10045*	Lemma for dfac5 10051. (Contributed by NM, 12-Apr-2004.)
⊢ (∃!𝑣 𝑣 ∈ (({𝑤} × 𝑤) ∩ 𝑦) ↔ ∃!𝑔(𝑔 ∈ 𝑤 ∧ ⟨𝑤, 𝑔⟩ ∈ 𝑦))
Theorem	dfac5lem2 10046*	Lemma for dfac5 10051. (Contributed by NM, 12-Apr-2004.)
⊢ 𝐴 = {𝑢 ∣ (𝑢 ≠ ∅ ∧ ∃𝑡 ∈ ℎ 𝑢 = ({𝑡} × 𝑡))}    ⇒   ⊢ (⟨𝑤, 𝑔⟩ ∈ ∪ 𝐴 ↔ (𝑤 ∈ ℎ ∧ 𝑔 ∈ 𝑤))
Theorem	dfac5lem3 10047*	Lemma for dfac5 10051. (Contributed by NM, 12-Apr-2004.)
⊢ 𝐴 = {𝑢 ∣ (𝑢 ≠ ∅ ∧ ∃𝑡 ∈ ℎ 𝑢 = ({𝑡} × 𝑡))}    ⇒   ⊢ (({𝑤} × 𝑤) ∈ 𝐴 ↔ (𝑤 ≠ ∅ ∧ 𝑤 ∈ ℎ))
Theorem	dfac5lem4 10048*	Lemma for dfac5 10051. (Contributed by NM, 11-Apr-2004.) Avoid ax-11 2163. (Revised by BTernaryTau, 23-Jun-2025.)
⊢ 𝐴 = {𝑢 ∣ (𝑢 ≠ ∅ ∧ ∃𝑡 ∈ ℎ 𝑢 = ({𝑡} × 𝑡))}    &   ⊢ (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	dfac5lem5 10049*	Lemma for dfac5 10051. (Contributed by NM, 12-Apr-2004.)
⊢ 𝐴 = {𝑢 ∣ (𝑢 ≠ ∅ ∧ ∃𝑡 ∈ ℎ 𝑢 = ({𝑡} × 𝑡))}    &   ⊢ (𝜑 ↔ ∀𝑥((∀𝑧 ∈ 𝑥 𝑧 ≠ ∅ ∧ ∀𝑧 ∈ 𝑥 ∀𝑤 ∈ 𝑥 (𝑧 ≠ 𝑤 → (𝑧 ∩ 𝑤) = ∅)) → ∃𝑦∀𝑧 ∈ 𝑥 ∃!𝑣 𝑣 ∈ (𝑧 ∩ 𝑦)))    &   ⊢ 𝐵 = (∪ 𝐴 ∩ 𝑦)    ⇒   ⊢ (𝜑 → ∃𝑓∀𝑤 ∈ ℎ (𝑤 ≠ ∅ → (𝑓‘𝑤) ∈ 𝑤))
Theorem	dfac5lem4OLD 10050*	Obsolete version of dfac5lem4 10048 as of 23-Jun-2025. (Contributed by NM, 11-Apr-2004.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ 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	dfac5 10051*	Equivalence of two versions of the Axiom of Choice. 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. (Contributed by NM, 11-Apr-2004.) (Revised by Mario Carneiro, 17-May-2015.)
⊢ (CHOICE ↔ ∀𝑥((∀𝑧 ∈ 𝑥 𝑧 ≠ ∅ ∧ ∀𝑧 ∈ 𝑥 ∀𝑤 ∈ 𝑥 (𝑧 ≠ 𝑤 → (𝑧 ∩ 𝑤) = ∅)) → ∃𝑦∀𝑧 ∈ 𝑥 ∃!𝑣 𝑣 ∈ (𝑧 ∩ 𝑦)))
Theorem	dfac2a 10052*	Our Axiom of Choice (in the form of ac3 10384) implies the Axiom of Choice (first form) of [Enderton] p. 49. The proof uses neither AC nor the Axiom of Regularity. See dfac2b 10053 for the converse (which does use the Axiom of Regularity). (Contributed by NM, 5-Apr-2004.) (Revised by Mario Carneiro, 26-Jun-2015.)
⊢ (∀𝑥∃𝑦∀𝑧 ∈ 𝑥 (𝑧 ≠ ∅ → ∃!𝑤 ∈ 𝑧 ∃𝑣 ∈ 𝑦 (𝑧 ∈ 𝑣 ∧ 𝑤 ∈ 𝑣)) → CHOICE)
Theorem	dfac2b 10053*	Axiom of Choice (first form) of [Enderton] p. 49 implies our Axiom of Choice (in the form of ac3 10384). The proof does not make use of AC. Note that the Axiom of Regularity is used by the proof. Specifically, elneq 9515 and preleq 9537 that are referenced in the proof each make use of Regularity for their derivations. (The reverse implication can be derived without using Regularity; see dfac2a 10052.) (Contributed by NM, 5-Apr-2004.) (Revised by Mario Carneiro, 26-Jun-2015.) (Revised by AV, 16-Jun-2022.)
⊢ (CHOICE → ∀𝑥∃𝑦∀𝑧 ∈ 𝑥 (𝑧 ≠ ∅ → ∃!𝑤 ∈ 𝑧 ∃𝑣 ∈ 𝑦 (𝑧 ∈ 𝑣 ∧ 𝑤 ∈ 𝑣)))
Theorem	dfac2 10054*	Axiom of Choice (first form) of [Enderton] p. 49 corresponds to our Axiom of Choice (in the form of ac3 10384). The proof does not make use of AC, but the Axiom of Regularity is used (by applying dfac2b 10053). (Contributed by NM, 5-Apr-2004.) (Revised by Mario Carneiro, 26-Jun-2015.) (Revised by AV, 16-Jun-2022.)
⊢ (CHOICE ↔ ∀𝑥∃𝑦∀𝑧 ∈ 𝑥 (𝑧 ≠ ∅ → ∃!𝑤 ∈ 𝑧 ∃𝑣 ∈ 𝑦 (𝑧 ∈ 𝑣 ∧ 𝑤 ∈ 𝑣)))
Theorem	dfac7 10055*	Equivalence of the Axiom of Choice (first form) of [Enderton] p. 49 and our Axiom of Choice (in the form of ac2 10383). The proof does not depend on AC but does depend on the Axiom of Regularity. (Contributed by Mario Carneiro, 17-May-2015.)
⊢ (CHOICE ↔ ∀𝑥∃𝑦∀𝑧 ∈ 𝑥 ∀𝑤 ∈ 𝑧 ∃!𝑣 ∈ 𝑧 ∃𝑢 ∈ 𝑦 (𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢))
Theorem	dfac0 10056*	Equivalence of two versions of the Axiom of Choice. The proof uses the Axiom of Regularity. The right-hand side is our original ax-ac 10381. (Contributed by Mario Carneiro, 17-May-2015.)
⊢ (CHOICE ↔ ∀𝑥∃𝑦∀𝑧∀𝑤((𝑧 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥) → ∃𝑣∀𝑢(∃𝑡((𝑢 ∈ 𝑤 ∧ 𝑤 ∈ 𝑡) ∧ (𝑢 ∈ 𝑡 ∧ 𝑡 ∈ 𝑦)) ↔ 𝑢 = 𝑣)))
Theorem	dfac1 10057*	Equivalence of two versions of the Axiom of Choice ax-ac 10381. The proof uses the Axiom of Regularity. The right-hand side expresses our AC with the fewest number of different variables. (Contributed by Mario Carneiro, 17-May-2015.)
⊢ (CHOICE ↔ ∀𝑥∃𝑦∀𝑧∀𝑤((𝑧 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥) → ∃𝑥∀𝑧(∃𝑥((𝑧 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥) ∧ (𝑧 ∈ 𝑥 ∧ 𝑥 ∈ 𝑦)) ↔ 𝑧 = 𝑥)))
Theorem	dfac8 10058*	A proof of the equivalency of the well-ordering theorem weth 10417 and the axiom of choice ac7 10395. (Contributed by Mario Carneiro, 5-Jan-2013.)
⊢ (CHOICE ↔ ∀𝑥∃𝑟 𝑟 We 𝑥)
Theorem	dfac9 10059*	Equivalence of the axiom of choice with a statement related to ac9 10405; definition AC3 of [Schechter] p. 139. (Contributed by Stefan O'Rear, 22-Feb-2015.)
⊢ (CHOICE ↔ ∀𝑓((Fun 𝑓 ∧ ∅ ∉ ran 𝑓) → X𝑥 ∈ dom 𝑓(𝑓‘𝑥) ≠ ∅))
Theorem	dfac10 10060	Axiom of Choice equivalent: the cardinality function measures every set. (Contributed by Mario Carneiro, 6-May-2015.)
⊢ (CHOICE ↔ dom card = V)
Theorem	dfac10c 10061*	Axiom of Choice equivalent: every set is equinumerous to an ordinal. (Contributed by Stefan O'Rear, 17-Jan-2015.)
⊢ (CHOICE ↔ ∀𝑥∃𝑦 ∈ On 𝑦 ≈ 𝑥)
Theorem	dfac10b 10062	Axiom of Choice equivalent: every set is equinumerous to an ordinal (quantifier-free short cryptic version alluded to in df-ac 10038). (Contributed by Stefan O'Rear, 17-Jan-2015.)
⊢ (CHOICE ↔ ( ≈ “ On) = V)
Theorem	acacni 10063	A choice equivalent: every set has choice sets of every length. (Contributed by Mario Carneiro, 31-Aug-2015.)
⊢ ((CHOICE ∧ 𝐴 ∈ 𝑉) → AC 𝐴 = V)
Theorem	dfacacn 10064	A choice equivalent: every set has choice sets of every length. (Contributed by Mario Carneiro, 31-Aug-2015.)
⊢ (CHOICE ↔ ∀𝑥AC 𝑥 = V)
Theorem	dfac13 10065	The axiom of choice holds iff every set has choice sequences as long as itself. (Contributed by Mario Carneiro, 3-Sep-2015.)
⊢ (CHOICE ↔ ∀𝑥 𝑥 ∈ AC 𝑥)
Theorem	dfac12lem1 10066*	Lemma for dfac12 10072. (Contributed by Mario Carneiro, 29-May-2015.)
⊢ (𝜑 → 𝐴 ∈ On)    &   ⊢ (𝜑 → 𝐹:𝒫 (har‘(𝑅1‘𝐴))–1-1→On)    &   ⊢ 𝐺 = recs((𝑥 ∈ V ↦ (𝑦 ∈ (𝑅1‘dom 𝑥) ↦ if(dom 𝑥 = ∪ dom 𝑥, ((suc ∪ ran ∪ ran 𝑥 ·o (rank‘𝑦)) +o ((𝑥‘suc (rank‘𝑦))‘𝑦)), (𝐹‘((◡OrdIso( E , ran (𝑥‘∪ dom 𝑥)) ∘ (𝑥‘∪ dom 𝑥)) “ 𝑦))))))    &   ⊢ (𝜑 → 𝐶 ∈ On)    &   ⊢ 𝐻 = (◡OrdIso( E , ran (𝐺‘∪ 𝐶)) ∘ (𝐺‘∪ 𝐶))    ⇒   ⊢ (𝜑 → (𝐺‘𝐶) = (𝑦 ∈ (𝑅1‘𝐶) ↦ if(𝐶 = ∪ 𝐶, ((suc ∪ ran ∪ (𝐺 “ 𝐶) ·o (rank‘𝑦)) +o ((𝐺‘suc (rank‘𝑦))‘𝑦)), (𝐹‘(𝐻 “ 𝑦)))))
Theorem	dfac12lem2 10067*	Lemma for dfac12 10072. (Contributed by Mario Carneiro, 29-May-2015.)
⊢ (𝜑 → 𝐴 ∈ On)    &   ⊢ (𝜑 → 𝐹:𝒫 (har‘(𝑅1‘𝐴))–1-1→On)    &   ⊢ 𝐺 = recs((𝑥 ∈ V ↦ (𝑦 ∈ (𝑅1‘dom 𝑥) ↦ if(dom 𝑥 = ∪ dom 𝑥, ((suc ∪ ran ∪ ran 𝑥 ·o (rank‘𝑦)) +o ((𝑥‘suc (rank‘𝑦))‘𝑦)), (𝐹‘((◡OrdIso( E , ran (𝑥‘∪ dom 𝑥)) ∘ (𝑥‘∪ dom 𝑥)) “ 𝑦))))))    &   ⊢ (𝜑 → 𝐶 ∈ On)    &   ⊢ 𝐻 = (◡OrdIso( E , ran (𝐺‘∪ 𝐶)) ∘ (𝐺‘∪ 𝐶))    &   ⊢ (𝜑 → 𝐶 ⊆ 𝐴)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝐶 (𝐺‘𝑧):(𝑅1‘𝑧)–1-1→On)    ⇒   ⊢ (𝜑 → (𝐺‘𝐶):(𝑅1‘𝐶)–1-1→On)
Theorem	dfac12lem3 10068*	Lemma for dfac12 10072. (Contributed by Mario Carneiro, 29-May-2015.)
⊢ (𝜑 → 𝐴 ∈ On)    &   ⊢ (𝜑 → 𝐹:𝒫 (har‘(𝑅1‘𝐴))–1-1→On)    &   ⊢ 𝐺 = recs((𝑥 ∈ V ↦ (𝑦 ∈ (𝑅1‘dom 𝑥) ↦ if(dom 𝑥 = ∪ dom 𝑥, ((suc ∪ ran ∪ ran 𝑥 ·o (rank‘𝑦)) +o ((𝑥‘suc (rank‘𝑦))‘𝑦)), (𝐹‘((◡OrdIso( E , ran (𝑥‘∪ dom 𝑥)) ∘ (𝑥‘∪ dom 𝑥)) “ 𝑦))))))    ⇒   ⊢ (𝜑 → (𝑅1‘𝐴) ∈ dom card)
Theorem	dfac12r 10069	The axiom of choice holds iff every ordinal has a well-orderable powerset. This version of dfac12 10072 does not assume the Axiom of Regularity. (Contributed by Mario Carneiro, 29-May-2015.)
⊢ (∀𝑥 ∈ On 𝒫 𝑥 ∈ dom card ↔ ∪ (𝑅1 “ On) ⊆ dom card)
Theorem	dfac12k 10070*	Equivalence of dfac12 10072 and dfac12a 10071, without using Regularity. (Contributed by Mario Carneiro, 21-May-2015.)
⊢ (∀𝑥 ∈ On 𝒫 𝑥 ∈ dom card ↔ ∀𝑦 ∈ On 𝒫 (ℵ‘𝑦) ∈ dom card)
Theorem	dfac12a 10071	The axiom of choice holds iff every ordinal has a well-orderable powerset. (Contributed by Mario Carneiro, 29-May-2015.)
⊢ (CHOICE ↔ ∀𝑥 ∈ On 𝒫 𝑥 ∈ dom card)
Theorem	dfac12 10072	The axiom of choice holds iff every aleph has a well-orderable powerset. (Contributed by Mario Carneiro, 21-May-2015.)
⊢ (CHOICE ↔ ∀𝑥 ∈ On 𝒫 (ℵ‘𝑥) ∈ dom card)
Theorem	kmlem1 10073*	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, 1 => 2. (Contributed by NM, 5-Apr-2004.)
⊢ (∀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 10074*	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4. (Contributed by NM, 25-Mar-2004.)
⊢ (∃𝑦∀𝑧 ∈ 𝑥 (𝜑 → ∃!𝑤 𝑤 ∈ (𝑧 ∩ 𝑦)) ↔ ∃𝑦(¬ 𝑦 ∈ 𝑥 ∧ ∀𝑧 ∈ 𝑥 (𝜑 → ∃!𝑤 𝑤 ∈ (𝑧 ∩ 𝑦))))
Theorem	kmlem3 10075*	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. (Contributed by NM, 25-Mar-2004.)
⊢ ((𝑧 ∖ ∪ (𝑥 ∖ {𝑧})) ≠ ∅ ↔ ∃𝑣 ∈ 𝑧 ∀𝑤 ∈ 𝑥 (𝑧 ≠ 𝑤 → ¬ 𝑣 ∈ (𝑧 ∩ 𝑤)))
Theorem	kmlem4 10076*	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4. (Contributed by NM, 26-Mar-2004.)
⊢ ((𝑤 ∈ 𝑥 ∧ 𝑧 ≠ 𝑤) → ((𝑧 ∖ ∪ (𝑥 ∖ {𝑧})) ∩ 𝑤) = ∅)
Theorem	kmlem5 10077*	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4. (Contributed by NM, 25-Mar-2004.)
⊢ ((𝑤 ∈ 𝑥 ∧ 𝑧 ≠ 𝑤) → ((𝑧 ∖ ∪ (𝑥 ∖ {𝑧})) ∩ (𝑤 ∖ ∪ (𝑥 ∖ {𝑤}))) = ∅)
Theorem	kmlem6 10078*	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 4 => 1. (Contributed by NM, 26-Mar-2004.)
⊢ ((∀𝑧 ∈ 𝑥 𝑧 ≠ ∅ ∧ ∀𝑧 ∈ 𝑥 ∀𝑤 ∈ 𝑥 (𝜑 → 𝐴 = ∅)) → ∀𝑧 ∈ 𝑥 ∃𝑣 ∈ 𝑧 ∀𝑤 ∈ 𝑥 (𝜑 → ¬ 𝑣 ∈ 𝐴))
Theorem	kmlem7 10079*	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 4 => 1. (Contributed by NM, 26-Mar-2004.)
⊢ ((∀𝑧 ∈ 𝑥 𝑧 ≠ ∅ ∧ ∀𝑧 ∈ 𝑥 ∀𝑤 ∈ 𝑥 (𝑧 ≠ 𝑤 → (𝑧 ∩ 𝑤) = ∅)) → ¬ ∃𝑧 ∈ 𝑥 ∀𝑣 ∈ 𝑧 ∃𝑤 ∈ 𝑥 (𝑧 ≠ 𝑤 ∧ 𝑣 ∈ (𝑧 ∩ 𝑤)))
Theorem	kmlem8 10080*	Lemma for 5-quantifier AC of Kurt Maes, Th. 4 1 <=> 4. (Contributed by NM, 4-Apr-2004.)
⊢ ((¬ ∃𝑧 ∈ 𝑢 ∀𝑤 ∈ 𝑧 𝜓 → ∃𝑦∀𝑧 ∈ 𝑢 (𝑧 ≠ ∅ → ∃!𝑤 𝑤 ∈ (𝑧 ∩ 𝑦))) ↔ (∃𝑧 ∈ 𝑢 ∀𝑤 ∈ 𝑧 𝜓 ∨ ∃𝑦(¬ 𝑦 ∈ 𝑢 ∧ ∀𝑧 ∈ 𝑢 ∃!𝑤 𝑤 ∈ (𝑧 ∩ 𝑦))))
Theorem	kmlem9 10081*	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4. (Contributed by NM, 25-Mar-2004.)
⊢ 𝐴 = {𝑢 ∣ ∃𝑡 ∈ 𝑥 𝑢 = (𝑡 ∖ ∪ (𝑥 ∖ {𝑡}))}    ⇒   ⊢ ∀𝑧 ∈ 𝐴 ∀𝑤 ∈ 𝐴 (𝑧 ≠ 𝑤 → (𝑧 ∩ 𝑤) = ∅)
Theorem	kmlem10 10082*	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4. (Contributed by NM, 25-Mar-2004.)
⊢ 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	kmlem11 10083*	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4. (Contributed by NM, 26-Mar-2004.)
⊢ 𝐴 = {𝑢 ∣ ∃𝑡 ∈ 𝑥 𝑢 = (𝑡 ∖ ∪ (𝑥 ∖ {𝑡}))}    ⇒   ⊢ (𝑧 ∈ 𝑥 → (𝑧 ∩ ∪ 𝐴) = (𝑧 ∖ ∪ (𝑥 ∖ {𝑧})))
Theorem	kmlem12 10084*	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4. (Contributed by NM, 27-Mar-2004.)
⊢ 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	kmlem13 10085*	Lemma for 5-quantifier AC of Kurt Maes, Th. 4 1 <=> 4. (Contributed by NM, 5-Apr-2004.)
⊢ 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	kmlem14 10086*	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 5 <=> 4. (Contributed by NM, 4-Apr-2004.)
⊢ (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 10087*	Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 5 <=> 4. (Contributed by NM, 4-Apr-2004.)
⊢ (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 10088*	Lemma for 5-quantifier AC of Kurt Maes, Th. 4 5 <=> 4. (Contributed by NM, 4-Apr-2004.)
⊢ (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(¬ 𝑦 ∈ 𝑥 ∧ 𝜒)) ↔ ∃𝑦∀𝑧∃𝑣∀𝑢((𝑦 ∈ 𝑥 ∧ 𝜑) ∨ (¬ 𝑦 ∈ 𝑥 ∧ 𝜓)))
Theorem	dfackm 10089*	Equivalence of the Axiom of Choice and Maes' AC ackm 10387. The proof consists of lemmas kmlem1 10073 through kmlem16 10088 and this final theorem. AC is not used for the proof. Note: bypassing the first step (i.e., replacing dfac5 10051 with biid 261) establishes the AC equivalence shown by Maes' writeup. The left-hand-side AC shown here was chosen because it is shorter to display. (Contributed by NM, 13-Apr-2004.) (Revised by Mario Carneiro, 17-May-2015.)
⊢ (CHOICE ↔ ∀𝑥∃𝑦∀𝑧∃𝑣∀𝑢((𝑦 ∈ 𝑥 ∧ (𝑧 ∈ 𝑦 → ((𝑣 ∈ 𝑥 ∧ ¬ 𝑦 = 𝑣) ∧ 𝑧 ∈ 𝑣))) ∨ (¬ 𝑦 ∈ 𝑥 ∧ (𝑧 ∈ 𝑥 → ((𝑣 ∈ 𝑧 ∧ 𝑣 ∈ 𝑦) ∧ ((𝑢 ∈ 𝑧 ∧ 𝑢 ∈ 𝑦) → 𝑢 = 𝑣))))))
2.6.14  Cardinal number arithmetic For cardinal arithmetic, we follow [Mendelson] p. 258. Rather than defining operations restricted to cardinal numbers, we use disjoint union df-dju 9825 (⊔) for cardinal addition, Cartesian product df-xp 5637 (×) for cardinal multiplication, and set exponentiation df-map 8775 (↑m) for cardinal exponentiation. Equinumerosity and dominance serve the roles of equality and ordering. If we wanted to, we could easily convert our theorems to actual cardinal number operations via carden 10473, carddom 10476, and cardsdom 10477. The advantage of Mendelson's approach is that we can directly use many equinumerosity theorems that we already have available.
Theorem	undjudom 10090	Cardinal addition dominates union. (Contributed by NM, 28-Sep-2004.) (Revised by Jim Kingdon, 15-Aug-2023.)
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (𝐴 ∪ 𝐵) ≼ (𝐴 ⊔ 𝐵))
Theorem	endjudisj 10091	Equinumerosity of a disjoint union and a union of two disjoint sets. (Contributed by NM, 5-Apr-2007.)
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ (𝐴 ∩ 𝐵) = ∅) → (𝐴 ⊔ 𝐵) ≈ (𝐴 ∪ 𝐵))
Theorem	djuen 10092	Disjoint unions of equinumerous sets are equinumerous. (Contributed by NM, 28-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
⊢ ((𝐴 ≈ 𝐵 ∧ 𝐶 ≈ 𝐷) → (𝐴 ⊔ 𝐶) ≈ (𝐵 ⊔ 𝐷))
Theorem	djuenun 10093	Disjoint union is equinumerous to union for disjoint sets. (Contributed by Mario Carneiro, 29-Apr-2015.) (Revised by Jim Kingdon, 19-Aug-2023.)
⊢ ((𝐴 ≈ 𝐵 ∧ 𝐶 ≈ 𝐷 ∧ (𝐵 ∩ 𝐷) = ∅) → (𝐴 ⊔ 𝐶) ≈ (𝐵 ∪ 𝐷))
Theorem	dju1en 10094	Cardinal addition with cardinal one (which is the same as ordinal one). Used in proof of Theorem 6J of [Enderton] p. 143. (Contributed by NM, 28-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
⊢ ((𝐴 ∈ 𝑉 ∧ ¬ 𝐴 ∈ 𝐴) → (𝐴 ⊔ 1o) ≈ suc 𝐴)
Theorem	dju1dif 10095	Adding and subtracting one gives back the original cardinality. Similar to pncan 11399 for cardinalities. (Contributed by Mario Carneiro, 18-May-2015.) (Revised by Jim Kingdon, 20-Aug-2023.)
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ (𝐴 ⊔ 1o)) → ((𝐴 ⊔ 1o) ∖ {𝐵}) ≈ 𝐴)
Theorem	dju1p1e2 10096	1+1=2 for cardinal number addition, derived from pm54.43 9925 as promised. Theorem *110.643 of Principia Mathematica, vol. II, p. 86, which adds the remark, "The above proposition is occasionally useful." Whitehead and Russell define cardinal addition on collections of all sets equinumerous to 1 and 2 (which for us are proper classes unless we restrict them as in karden 9819), but after applying definitions, our theorem is equivalent. Because we use a disjoint union for cardinal addition (as explained in the comment at the top of this section), we use ≈ instead of =. See dju1p1e2ALT 10097 for a shorter proof that doesn't use pm54.43 9925. (Contributed by NM, 5-Apr-2007.) (Proof modification is discouraged.)
⊢ (1o ⊔ 1o) ≈ 2o
Theorem	dju1p1e2ALT 10097	Alternate proof of dju1p1e2 10096. (Contributed by Mario Carneiro, 29-Apr-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ (1o ⊔ 1o) ≈ 2o
Theorem	dju0en 10098	Cardinal addition with cardinal zero (the empty set). Part (a1) of proof of Theorem 6J of [Enderton] p. 143. (Contributed by NM, 27-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
⊢ (𝐴 ∈ 𝑉 → (𝐴 ⊔ ∅) ≈ 𝐴)
Theorem	xp2dju 10099	Two times a cardinal number. Exercise 4.56(g) of [Mendelson] p. 258. (Contributed by NM, 27-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
⊢ (2o × 𝐴) = (𝐴 ⊔ 𝐴)
Theorem	djucomen 10100	Commutative law for cardinal addition. Exercise 4.56(c) of [Mendelson] p. 258. (Contributed by NM, 24-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (𝐴 ⊔ 𝐵) ≈ (𝐵 ⊔ 𝐴))