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	alephfplem1 10001	Lemma for alephfp 10005. (Contributed by NM, 6-Nov-2004.)
⊢ 𝐻 = (rec(ℵ, ω) ↾ ω)    ⇒   ⊢ (𝐻‘∅) ∈ ran ℵ
Theorem	alephfplem2 10002*	Lemma for alephfp 10005. (Contributed by NM, 6-Nov-2004.)
⊢ 𝐻 = (rec(ℵ, ω) ↾ ω)    ⇒   ⊢ (𝑤 ∈ ω → (𝐻‘suc 𝑤) = (ℵ‘(𝐻‘𝑤)))
Theorem	alephfplem3 10003*	Lemma for alephfp 10005. (Contributed by NM, 6-Nov-2004.)
⊢ 𝐻 = (rec(ℵ, ω) ↾ ω)    ⇒   ⊢ (𝑣 ∈ ω → (𝐻‘𝑣) ∈ ran ℵ)
Theorem	alephfplem4 10004	Lemma for alephfp 10005. (Contributed by NM, 5-Nov-2004.)
⊢ 𝐻 = (rec(ℵ, ω) ↾ ω)    ⇒   ⊢ ∪ (𝐻 “ ω) ∈ ran ℵ
Theorem	alephfp 10005	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 10006 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 10006	The aleph function has at least one fixed point. Proposition 11.18 of [TakeutiZaring] p. 104. See alephfp 10005 for an actual example of a fixed point. Compare the inequality alephle 9985 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 10007*	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 10008	The power set of an aleph is not strictly dominated by the successor aleph. (The Generalized Continuum Hypothesis says they are equinumerous, see gch3 10573 or gchaleph2 10569.) The transposed form alephsucpw 10467 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 10009	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 10010*	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 10011	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 10012	Wff for an abbreviation of the axiom of choice.
wff CHOICE
Definition	df-ac 10013*	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 10356 as our definition, because the equivalence to more standard forms (dfac2 10029) 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 10356 itself as dfac0 10031. (Contributed by Mario Carneiro, 22-Feb-2015.)
⊢ (CHOICE ↔ ∀𝑥∃𝑓(𝑓 ⊆ 𝑥 ∧ 𝑓 Fn dom 𝑥))
Theorem	aceq1 10014*	Equivalence of two versions of the Axiom of Choice ax-ac 10356. 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 10015*	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 10356. (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 10016*	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 10017*	Lemma for dfac3 10018. (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 10018*	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 10019*	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 10020*	Lemma for dfac5 10026. (Contributed by NM, 12-Apr-2004.)
⊢ (∃!𝑣 𝑣 ∈ (({𝑤} × 𝑤) ∩ 𝑦) ↔ ∃!𝑔(𝑔 ∈ 𝑤 ∧ ⟨𝑤, 𝑔⟩ ∈ 𝑦))
Theorem	dfac5lem2 10021*	Lemma for dfac5 10026. (Contributed by NM, 12-Apr-2004.)
⊢ 𝐴 = {𝑢 ∣ (𝑢 ≠ ∅ ∧ ∃𝑡 ∈ ℎ 𝑢 = ({𝑡} × 𝑡))}    ⇒   ⊢ (⟨𝑤, 𝑔⟩ ∈ ∪ 𝐴 ↔ (𝑤 ∈ ℎ ∧ 𝑔 ∈ 𝑤))
Theorem	dfac5lem3 10022*	Lemma for dfac5 10026. (Contributed by NM, 12-Apr-2004.)
⊢ 𝐴 = {𝑢 ∣ (𝑢 ≠ ∅ ∧ ∃𝑡 ∈ ℎ 𝑢 = ({𝑡} × 𝑡))}    ⇒   ⊢ (({𝑤} × 𝑤) ∈ 𝐴 ↔ (𝑤 ≠ ∅ ∧ 𝑤 ∈ ℎ))
Theorem	dfac5lem4 10023*	Lemma for dfac5 10026. (Contributed by NM, 11-Apr-2004.) Avoid ax-11 2160. (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 10024*	Lemma for dfac5 10026. (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 10025*	Obsolete version of dfac5lem4 10023 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 10026*	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 10027*	Our Axiom of Choice (in the form of ac3 10359) implies the Axiom of Choice (first form) of [Enderton] p. 49. The proof uses neither AC nor the Axiom of Regularity. See dfac2b 10028 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 10028*	Axiom of Choice (first form) of [Enderton] p. 49 implies our Axiom of Choice (in the form of ac3 10359). The proof does not make use of AC. Note that the Axiom of Regularity is used by the proof. Specifically, elneq 9492 and preleq 9512 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 10027.) (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 10029*	Axiom of Choice (first form) of [Enderton] p. 49 corresponds to our Axiom of Choice (in the form of ac3 10359). The proof does not make use of AC, but the Axiom of Regularity is used (by applying dfac2b 10028). (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 10030*	Equivalence of the Axiom of Choice (first form) of [Enderton] p. 49 and our Axiom of Choice (in the form of ac2 10358). 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 10031*	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 10356. (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 10032*	Equivalence of two versions of the Axiom of Choice ax-ac 10356. 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 10033*	A proof of the equivalency of the well-ordering theorem weth 10392 and the axiom of choice ac7 10370. (Contributed by Mario Carneiro, 5-Jan-2013.)
⊢ (CHOICE ↔ ∀𝑥∃𝑟 𝑟 We 𝑥)
Theorem	dfac9 10034*	Equivalence of the axiom of choice with a statement related to ac9 10380; 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 10035	Axiom of Choice equivalent: the cardinality function measures every set. (Contributed by Mario Carneiro, 6-May-2015.)
⊢ (CHOICE ↔ dom card = V)
Theorem	dfac10c 10036*	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 10037	Axiom of Choice equivalent: every set is equinumerous to an ordinal (quantifier-free short cryptic version alluded to in df-ac 10013). (Contributed by Stefan O'Rear, 17-Jan-2015.)
⊢ (CHOICE ↔ ( ≈ “ On) = V)
Theorem	acacni 10038	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 10039	A choice equivalent: every set has choice sets of every length. (Contributed by Mario Carneiro, 31-Aug-2015.)
⊢ (CHOICE ↔ ∀𝑥AC 𝑥 = V)
Theorem	dfac13 10040	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 10041*	Lemma for dfac12 10047. (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 10042*	Lemma for dfac12 10047. (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 10043*	Lemma for dfac12 10047. (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 10044	The axiom of choice holds iff every ordinal has a well-orderable powerset. This version of dfac12 10047 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 10045*	Equivalence of dfac12 10047 and dfac12a 10046, 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 10046	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 10047	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 10048*	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 10049*	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 10050*	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 10051*	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 10052*	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 10053*	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 10054*	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 10055*	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 10056*	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 10057*	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 10058*	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 10059*	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 10060*	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 10061*	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 10062*	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 10063*	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 10064*	Equivalence of the Axiom of Choice and Maes' AC ackm 10362. The proof consists of lemmas kmlem1 10048 through kmlem16 10063 and this final theorem. AC is not used for the proof. Note: bypassing the first step (i.e., replacing dfac5 10026 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 9800 (⊔) for cardinal addition, Cartesian product df-xp 5625 (×) for cardinal multiplication, and set exponentiation df-map 8758 (↑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 10448, carddom 10451, and cardsdom 10452. The advantage of Mendelson's approach is that we can directly use many equinumerosity theorems that we already have available.
Theorem	undjudom 10065	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 10066	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 10067	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 10068	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 10069	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 10070	Adding and subtracting one gives back the original cardinality. Similar to pncan 11372 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 10071	1+1=2 for cardinal number addition, derived from pm54.43 9900 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 9794), 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 10072 for a shorter proof that doesn't use pm54.43 9900. (Contributed by NM, 5-Apr-2007.) (Proof modification is discouraged.)
⊢ (1o ⊔ 1o) ≈ 2o
Theorem	dju1p1e2ALT 10072	Alternate proof of dju1p1e2 10071. (Contributed by Mario Carneiro, 29-Apr-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ (1o ⊔ 1o) ≈ 2o
Theorem	dju0en 10073	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 10074	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 10075	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))
Theorem	djuassen 10076	Associative law for cardinal addition. Exercise 4.56(c) of [Mendelson] p. 258. (Contributed by NM, 26-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝐶 ∈ 𝑋) → ((𝐴 ⊔ 𝐵) ⊔ 𝐶) ≈ (𝐴 ⊔ (𝐵 ⊔ 𝐶)))
Theorem	xpdjuen 10077	Cardinal multiplication distributes over cardinal addition. Theorem 6I(3) of [Enderton] p. 142. (Contributed by NM, 26-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝐶 ∈ 𝑋) → (𝐴 × (𝐵 ⊔ 𝐶)) ≈ ((𝐴 × 𝐵) ⊔ (𝐴 × 𝐶)))
Theorem	mapdjuen 10078	Sum of exponents law for cardinal arithmetic. Theorem 6I(4) of [Enderton] p. 142. (Contributed by NM, 27-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝐶 ∈ 𝑋) → (𝐴 ↑m (𝐵 ⊔ 𝐶)) ≈ ((𝐴 ↑m 𝐵) × (𝐴 ↑m 𝐶)))
Theorem	pwdjuen 10079	Sum of exponents law for cardinal arithmetic. (Contributed by Mario Carneiro, 15-May-2015.)
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → 𝒫 (𝐴 ⊔ 𝐵) ≈ (𝒫 𝐴 × 𝒫 𝐵))
Theorem	djudom1 10080	Ordering law for cardinal addition. Exercise 4.56(f) of [Mendelson] p. 258. (Contributed by NM, 28-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.) (Revised by Jim Kingdon, 1-Sep-2023.)
⊢ ((𝐴 ≼ 𝐵 ∧ 𝐶 ∈ 𝑉) → (𝐴 ⊔ 𝐶) ≼ (𝐵 ⊔ 𝐶))
Theorem	djudom2 10081	Ordering law for cardinal addition. Theorem 6L(a) of [Enderton] p. 149. (Contributed by NM, 28-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
⊢ ((𝐴 ≼ 𝐵 ∧ 𝐶 ∈ 𝑉) → (𝐶 ⊔ 𝐴) ≼ (𝐶 ⊔ 𝐵))
Theorem	djudoml 10082	A set is dominated by its disjoint union with another. (Contributed by NM, 28-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → 𝐴 ≼ (𝐴 ⊔ 𝐵))
Theorem	djuxpdom 10083	Cartesian product dominates disjoint union for sets with cardinality greater than 1. Similar to Proposition 10.36 of [TakeutiZaring] p. 93. (Contributed by Mario Carneiro, 18-May-2015.)
⊢ ((1o ≺ 𝐴 ∧ 1o ≺ 𝐵) → (𝐴 ⊔ 𝐵) ≼ (𝐴 × 𝐵))
Theorem	djufi 10084	The disjoint union of two finite sets is finite. (Contributed by NM, 22-Oct-2004.)
⊢ ((𝐴 ≺ ω ∧ 𝐵 ≺ ω) → (𝐴 ⊔ 𝐵) ≺ ω)
Theorem	cdainflem 10085	Any partition of omega into two pieces (which may be disjoint) contains an infinite subset. (Contributed by Mario Carneiro, 11-Feb-2013.)
⊢ ((𝐴 ∪ 𝐵) ≈ ω → (𝐴 ≈ ω ∨ 𝐵 ≈ ω))
Theorem	djuinf 10086	A set is infinite iff the cardinal sum with itself is infinite. (Contributed by NM, 22-Oct-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
⊢ (ω ≼ 𝐴 ↔ ω ≼ (𝐴 ⊔ 𝐴))
Theorem	infdju1 10087	An infinite set is equinumerous to itself added with one. (Contributed by Mario Carneiro, 15-May-2015.)
⊢ (ω ≼ 𝐴 → (𝐴 ⊔ 1o) ≈ 𝐴)
Theorem	pwdju1 10088	The sum of a powerset with itself is equipotent to the successor powerset. (Contributed by Mario Carneiro, 15-May-2015.)
⊢ (𝐴 ∈ 𝑉 → (𝒫 𝐴 ⊔ 𝒫 𝐴) ≈ 𝒫 (𝐴 ⊔ 1o))
Theorem	pwdjuidm 10089	If the natural numbers inject into 𝐴, then 𝒫 𝐴 is idempotent under cardinal sum. (Contributed by Mario Carneiro, 15-May-2015.)
⊢ (ω ≼ 𝐴 → (𝒫 𝐴 ⊔ 𝒫 𝐴) ≈ 𝒫 𝐴)
Theorem	djulepw 10090	If 𝐴 is idempotent under cardinal sum and 𝐵 is dominated by the power set of 𝐴, then so is the cardinal sum of 𝐴 and 𝐵. (Contributed by Mario Carneiro, 15-May-2015.)
⊢ (((𝐴 ⊔ 𝐴) ≈ 𝐴 ∧ 𝐵 ≼ 𝒫 𝐴) → (𝐴 ⊔ 𝐵) ≼ 𝒫 𝐴)
Theorem	onadju 10091	The cardinal and ordinal sums are always equinumerous. (Contributed by Mario Carneiro, 6-Feb-2013.) (Revised by Jim Kingdon, 7-Sep-2023.)
⊢ ((𝐴 ∈ On ∧ 𝐵 ∈ On) → (𝐴 +o 𝐵) ≈ (𝐴 ⊔ 𝐵))
Theorem	cardadju 10092	The cardinal sum is equinumerous to an ordinal sum of the cardinals. (Contributed by Mario Carneiro, 6-Feb-2013.) (Revised by Mario Carneiro, 28-Apr-2015.)
⊢ ((𝐴 ∈ dom card ∧ 𝐵 ∈ dom card) → (𝐴 ⊔ 𝐵) ≈ ((card‘𝐴) +o (card‘𝐵)))
Theorem	djunum 10093	The disjoint union of two numerable sets is numerable. (Contributed by Mario Carneiro, 29-Apr-2015.)
⊢ ((𝐴 ∈ dom card ∧ 𝐵 ∈ dom card) → (𝐴 ⊔ 𝐵) ∈ dom card)
Theorem	unnum 10094	The union of two numerable sets is numerable. (Contributed by Mario Carneiro, 29-Apr-2015.)
⊢ ((𝐴 ∈ dom card ∧ 𝐵 ∈ dom card) → (𝐴 ∪ 𝐵) ∈ dom card)
Theorem	nnadju 10095	The cardinal and ordinal sums of finite ordinals are equal. For a shorter proof using ax-rep 5219, see nnadjuALT 10096. (Contributed by Paul Chapman, 11-Apr-2009.) (Revised by Mario Carneiro, 6-Feb-2013.) Avoid ax-rep 5219. (Revised by BTernaryTau, 2-Jul-2024.)
⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (card‘(𝐴 ⊔ 𝐵)) = (𝐴 +o 𝐵))
Theorem	nnadjuALT 10096	Shorter proof of nnadju 10095 using ax-rep 5219. (Contributed by Paul Chapman, 11-Apr-2009.) (Revised by Mario Carneiro, 6-Feb-2013.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (card‘(𝐴 ⊔ 𝐵)) = (𝐴 +o 𝐵))
Theorem	ficardadju 10097	The disjoint union of finite sets is equinumerous to the ordinal sum of the cardinalities of those sets. (Contributed by BTernaryTau, 3-Jul-2024.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) → (𝐴 ⊔ 𝐵) ≈ ((card‘𝐴) +o (card‘𝐵)))
Theorem	ficardun 10098	The cardinality of the union of disjoint, finite sets is the ordinal sum of their cardinalities. (Contributed by Paul Chapman, 5-Jun-2009.) (Proof shortened by Mario Carneiro, 28-Apr-2015.) Avoid ax-rep 5219. (Revised by BTernaryTau, 3-Jul-2024.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin ∧ (𝐴 ∩ 𝐵) = ∅) → (card‘(𝐴 ∪ 𝐵)) = ((card‘𝐴) +o (card‘𝐵)))
Theorem	ficardun2 10099	The cardinality of the union of finite sets is at most the ordinal sum of their cardinalities. (Contributed by Mario Carneiro, 5-Feb-2013.) Avoid ax-rep 5219. (Revised by BTernaryTau, 3-Jul-2024.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) → (card‘(𝐴 ∪ 𝐵)) ⊆ ((card‘𝐴) +o (card‘𝐵)))
Theorem	pwsdompw 10100*	Lemma for domtriom 10340. This is the equinumerosity version of the algebraic identity Σ𝑘 ∈ 𝑛(2↑𝑘) = (2↑𝑛) − 1. (Contributed by Mario Carneiro, 7-Feb-2013.)
⊢ ((𝑛 ∈ ω ∧ ∀𝑘 ∈ suc 𝑛(𝐵‘𝑘) ≈ 𝒫 𝑘) → ∪ 𝑘 ∈ 𝑛 (𝐵‘𝑘) ≺ (𝐵‘𝑛))