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Theorem List for Metamath Proof Explorer - 9801-9900   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremiunfictbso 9801 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)
 
2.6.12  Axiom of Choice equivalents
 
Syntaxwac 9802 Wff for an abbreviation of the axiom of choice.
wff CHOICE
 
Definitiondf-ac 9803* 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 10146 as our definition, because the equivalence to more standard forms (dfac2 9818) 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 10146 itself as dfac0 9820. (Contributed by Mario Carneiro, 22-Feb-2015.)

(CHOICE ↔ ∀𝑥𝑓(𝑓𝑥𝑓 Fn dom 𝑥))
 
Theoremaceq1 9804* Equivalence of two versions of the Axiom of Choice ax-ac 10146. 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.)
(∃𝑦𝑧𝑥𝑤𝑧 ∃!𝑣𝑧𝑢𝑦 (𝑧𝑢𝑣𝑢) ↔ ∃𝑦𝑧𝑤((𝑧𝑤𝑤𝑥) → ∃𝑥𝑧(∃𝑥((𝑧𝑤𝑤𝑥) ∧ (𝑧𝑥𝑥𝑦)) ↔ 𝑧 = 𝑥)))
 
Theoremaceq0 9805* 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 10146. (Contributed by NM, 5-Apr-2004.)
(∃𝑦𝑧𝑥𝑤𝑧 ∃!𝑣𝑧𝑢𝑦 (𝑧𝑢𝑣𝑢) ↔ ∃𝑦𝑧𝑤((𝑧𝑤𝑤𝑥) → ∃𝑣𝑢(∃𝑡((𝑢𝑤𝑤𝑡) ∧ (𝑢𝑡𝑡𝑦)) ↔ 𝑢 = 𝑣)))
 
Theoremaceq2 9806* 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.)
(∃𝑦𝑧𝑥𝑤𝑧 ∃!𝑣𝑧𝑢𝑦 (𝑧𝑢𝑣𝑢) ↔ ∃𝑦𝑧𝑥 (𝑧 ≠ ∅ → ∃!𝑤𝑧𝑣𝑦 (𝑧𝑣𝑤𝑣)))
 
Theoremaceq3lem 9807* Lemma for dfac3 9808. (Contributed by NM, 2-Apr-2004.) (Revised by Mario Carneiro, 26-Jun-2015.)
𝐹 = (𝑤 ∈ dom 𝑦 ↦ (𝑓‘{𝑢𝑤𝑦𝑢}))       (∀𝑥𝑓𝑧𝑥 (𝑧 ≠ ∅ → (𝑓𝑧) ∈ 𝑧) → ∃𝑓(𝑓𝑦𝑓 Fn dom 𝑦))
 
Theoremdfac3 9808* 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 ↔ ∀𝑥𝑓𝑧𝑥 (𝑧 ≠ ∅ → (𝑓𝑧) ∈ 𝑧))
 
Theoremdfac4 9809* 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 ↔ ∀𝑥𝑓(𝑓 Fn 𝑥 ∧ ∀𝑧𝑥 (𝑧 ≠ ∅ → (𝑓𝑧) ∈ 𝑧)))
 
Theoremdfac5lem1 9810* Lemma for dfac5 9815. (Contributed by NM, 12-Apr-2004.)
(∃!𝑣 𝑣 ∈ (({𝑤} × 𝑤) ∩ 𝑦) ↔ ∃!𝑔(𝑔𝑤 ∧ ⟨𝑤, 𝑔⟩ ∈ 𝑦))
 
Theoremdfac5lem2 9811* Lemma for dfac5 9815. (Contributed by NM, 12-Apr-2004.)
𝐴 = {𝑢 ∣ (𝑢 ≠ ∅ ∧ ∃𝑡 𝑢 = ({𝑡} × 𝑡))}       (⟨𝑤, 𝑔⟩ ∈ 𝐴 ↔ (𝑤𝑔𝑤))
 
Theoremdfac5lem3 9812* Lemma for dfac5 9815. (Contributed by NM, 12-Apr-2004.)
𝐴 = {𝑢 ∣ (𝑢 ≠ ∅ ∧ ∃𝑡 𝑢 = ({𝑡} × 𝑡))}       (({𝑤} × 𝑤) ∈ 𝐴 ↔ (𝑤 ≠ ∅ ∧ 𝑤))
 
Theoremdfac5lem4 9813* Lemma for dfac5 9815. (Contributed by NM, 11-Apr-2004.)
𝐴 = {𝑢 ∣ (𝑢 ≠ ∅ ∧ ∃𝑡 𝑢 = ({𝑡} × 𝑡))}    &   𝐵 = ( 𝐴𝑦)    &   (𝜑 ↔ ∀𝑥((∀𝑧𝑥 𝑧 ≠ ∅ ∧ ∀𝑧𝑥𝑤𝑥 (𝑧𝑤 → (𝑧𝑤) = ∅)) → ∃𝑦𝑧𝑥 ∃!𝑣 𝑣 ∈ (𝑧𝑦)))       (𝜑 → ∃𝑦𝑧𝐴 ∃!𝑣 𝑣 ∈ (𝑧𝑦))
 
Theoremdfac5lem5 9814* Lemma for dfac5 9815. (Contributed by NM, 12-Apr-2004.)
𝐴 = {𝑢 ∣ (𝑢 ≠ ∅ ∧ ∃𝑡 𝑢 = ({𝑡} × 𝑡))}    &   𝐵 = ( 𝐴𝑦)    &   (𝜑 ↔ ∀𝑥((∀𝑧𝑥 𝑧 ≠ ∅ ∧ ∀𝑧𝑥𝑤𝑥 (𝑧𝑤 → (𝑧𝑤) = ∅)) → ∃𝑦𝑧𝑥 ∃!𝑣 𝑣 ∈ (𝑧𝑦)))       (𝜑 → ∃𝑓𝑤 (𝑤 ≠ ∅ → (𝑓𝑤) ∈ 𝑤))
 
Theoremdfac5 9815* 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((∀𝑧𝑥 𝑧 ≠ ∅ ∧ ∀𝑧𝑥𝑤𝑥 (𝑧𝑤 → (𝑧𝑤) = ∅)) → ∃𝑦𝑧𝑥 ∃!𝑣 𝑣 ∈ (𝑧𝑦)))
 
Theoremdfac2a 9816* Our Axiom of Choice (in the form of ac3 10149) implies the Axiom of Choice (first form) of [Enderton] p. 49. The proof uses neither AC nor the Axiom of Regularity. See dfac2b 9817 for the converse (which does use the Axiom of Regularity). (Contributed by NM, 5-Apr-2004.) (Revised by Mario Carneiro, 26-Jun-2015.)
(∀𝑥𝑦𝑧𝑥 (𝑧 ≠ ∅ → ∃!𝑤𝑧𝑣𝑦 (𝑧𝑣𝑤𝑣)) → CHOICE)
 
Theoremdfac2b 9817* Axiom of Choice (first form) of [Enderton] p. 49 implies our Axiom of Choice (in the form of ac3 10149). The proof does not make use of AC. Note that the Axiom of Regularity is used by the proof. Specifically, elneq 9287 and preleq 9304 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 9816.) (Contributed by NM, 5-Apr-2004.) (Revised by Mario Carneiro, 26-Jun-2015.) (Revised by AV, 16-Jun-2022.)
(CHOICE → ∀𝑥𝑦𝑧𝑥 (𝑧 ≠ ∅ → ∃!𝑤𝑧𝑣𝑦 (𝑧𝑣𝑤𝑣)))
 
Theoremdfac2 9818* Axiom of Choice (first form) of [Enderton] p. 49 corresponds to our Axiom of Choice (in the form of ac3 10149). The proof does not make use of AC, but the Axiom of Regularity is used (by applying dfac2b 9817). (Contributed by NM, 5-Apr-2004.) (Revised by Mario Carneiro, 26-Jun-2015.) (Revised by AV, 16-Jun-2022.)
(CHOICE ↔ ∀𝑥𝑦𝑧𝑥 (𝑧 ≠ ∅ → ∃!𝑤𝑧𝑣𝑦 (𝑧𝑣𝑤𝑣)))
 
Theoremdfac7 9819* Equivalence of the Axiom of Choice (first form) of [Enderton] p. 49 and our Axiom of Choice (in the form of ac2 10148). The proof does not depend on AC but does depend on the Axiom of Regularity. (Contributed by Mario Carneiro, 17-May-2015.)
(CHOICE ↔ ∀𝑥𝑦𝑧𝑥𝑤𝑧 ∃!𝑣𝑧𝑢𝑦 (𝑧𝑢𝑣𝑢))
 
Theoremdfac0 9820* 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 10146. (Contributed by Mario Carneiro, 17-May-2015.)
(CHOICE ↔ ∀𝑥𝑦𝑧𝑤((𝑧𝑤𝑤𝑥) → ∃𝑣𝑢(∃𝑡((𝑢𝑤𝑤𝑡) ∧ (𝑢𝑡𝑡𝑦)) ↔ 𝑢 = 𝑣)))
 
Theoremdfac1 9821* Equivalence of two versions of the Axiom of Choice ax-ac 10146. 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 ↔ ∀𝑥𝑦𝑧𝑤((𝑧𝑤𝑤𝑥) → ∃𝑥𝑧(∃𝑥((𝑧𝑤𝑤𝑥) ∧ (𝑧𝑥𝑥𝑦)) ↔ 𝑧 = 𝑥)))
 
Theoremdfac8 9822* A proof of the equivalency of the well-ordering theorem weth 10182 and the axiom of choice ac7 10160. (Contributed by Mario Carneiro, 5-Jan-2013.)
(CHOICE ↔ ∀𝑥𝑟 𝑟 We 𝑥)
 
Theoremdfac9 9823* Equivalence of the axiom of choice with a statement related to ac9 10170; definition AC3 of [Schechter] p. 139. (Contributed by Stefan O'Rear, 22-Feb-2015.)
(CHOICE ↔ ∀𝑓((Fun 𝑓 ∧ ∅ ∉ ran 𝑓) → X𝑥 ∈ dom 𝑓(𝑓𝑥) ≠ ∅))
 
Theoremdfac10 9824 Axiom of Choice equivalent: the cardinality function measures every set. (Contributed by Mario Carneiro, 6-May-2015.)
(CHOICE ↔ dom card = V)
 
Theoremdfac10c 9825* Axiom of Choice equivalent: every set is equinumerous to an ordinal. (Contributed by Stefan O'Rear, 17-Jan-2015.)
(CHOICE ↔ ∀𝑥𝑦 ∈ On 𝑦𝑥)
 
Theoremdfac10b 9826 Axiom of Choice equivalent: every set is equinumerous to an ordinal (quantifier-free short cryptic version alluded to in df-ac 9803). (Contributed by Stefan O'Rear, 17-Jan-2015.)
(CHOICE ↔ ( ≈ “ On) = V)
 
Theoremacacni 9827 A choice equivalent: every set has choice sets of every length. (Contributed by Mario Carneiro, 31-Aug-2015.)
((CHOICE𝐴𝑉) → AC 𝐴 = V)
 
Theoremdfacacn 9828 A choice equivalent: every set has choice sets of every length. (Contributed by Mario Carneiro, 31-Aug-2015.)
(CHOICE ↔ ∀𝑥AC 𝑥 = V)
 
Theoremdfac13 9829 The axiom of choice holds iff every set has choice sequences as long as itself. (Contributed by Mario Carneiro, 3-Sep-2015.)
(CHOICE ↔ ∀𝑥 𝑥AC 𝑥)
 
Theoremdfac12lem1 9830* Lemma for dfac12 9836. (Contributed by Mario Carneiro, 29-May-2015.)
(𝜑𝐴 ∈ On)    &   (𝜑𝐹:𝒫 (har‘(𝑅1𝐴))–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))))))    &   (𝜑𝐶 ∈ On)    &   𝐻 = (OrdIso( E , ran (𝐺 𝐶)) ∘ (𝐺 𝐶))       (𝜑 → (𝐺𝐶) = (𝑦 ∈ (𝑅1𝐶) ↦ if(𝐶 = 𝐶, ((suc ran (𝐺𝐶) ·o (rank‘𝑦)) +o ((𝐺‘suc (rank‘𝑦))‘𝑦)), (𝐹‘(𝐻𝑦)))))
 
Theoremdfac12lem2 9831* Lemma for dfac12 9836. (Contributed by Mario Carneiro, 29-May-2015.)
(𝜑𝐴 ∈ On)    &   (𝜑𝐹:𝒫 (har‘(𝑅1𝐴))–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))))))    &   (𝜑𝐶 ∈ On)    &   𝐻 = (OrdIso( E , ran (𝐺 𝐶)) ∘ (𝐺 𝐶))    &   (𝜑𝐶𝐴)    &   (𝜑 → ∀𝑧𝐶 (𝐺𝑧):(𝑅1𝑧)–1-1→On)       (𝜑 → (𝐺𝐶):(𝑅1𝐶)–1-1→On)
 
Theoremdfac12lem3 9832* Lemma for dfac12 9836. (Contributed by Mario Carneiro, 29-May-2015.)
(𝜑𝐴 ∈ On)    &   (𝜑𝐹:𝒫 (har‘(𝑅1𝐴))–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 → (𝑅1𝐴) ∈ dom card)
 
Theoremdfac12r 9833 The axiom of choice holds iff every ordinal has a well-orderable powerset. This version of dfac12 9836 does not assume the Axiom of Regularity. (Contributed by Mario Carneiro, 29-May-2015.)
(∀𝑥 ∈ On 𝒫 𝑥 ∈ dom card ↔ (𝑅1 “ On) ⊆ dom card)
 
Theoremdfac12k 9834* Equivalence of dfac12 9836 and dfac12a 9835, without using Regularity. (Contributed by Mario Carneiro, 21-May-2015.)
(∀𝑥 ∈ On 𝒫 𝑥 ∈ dom card ↔ ∀𝑦 ∈ On 𝒫 (ℵ‘𝑦) ∈ dom card)
 
Theoremdfac12a 9835 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)
 
Theoremdfac12 9836 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)
 
Theoremkmlem1 9837* Lemma for 5-quantifier AC of Kurt Maes, Th. 4, 1 => 2. (Contributed by NM, 5-Apr-2004.)
(∀𝑥((∀𝑧𝑥 𝑧 ≠ ∅ ∧ ∀𝑧𝑥𝑤𝑥 𝜑) → ∃𝑦𝑧𝑥 𝜓) → ∀𝑥(∀𝑧𝑥𝑤𝑥 𝜑 → ∃𝑦𝑧𝑥 (𝑧 ≠ ∅ → 𝜓)))
 
Theoremkmlem2 9838* Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4. (Contributed by NM, 25-Mar-2004.)
(∃𝑦𝑧𝑥 (𝜑 → ∃!𝑤 𝑤 ∈ (𝑧𝑦)) ↔ ∃𝑦𝑦𝑥 ∧ ∀𝑧𝑥 (𝜑 → ∃!𝑤 𝑤 ∈ (𝑧𝑦))))
 
Theoremkmlem3 9839* 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})) ≠ ∅ ↔ ∃𝑣𝑧𝑤𝑥 (𝑧𝑤 → ¬ 𝑣 ∈ (𝑧𝑤)))
 
Theoremkmlem4 9840* Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4. (Contributed by NM, 26-Mar-2004.)
((𝑤𝑥𝑧𝑤) → ((𝑧 (𝑥 ∖ {𝑧})) ∩ 𝑤) = ∅)
 
Theoremkmlem5 9841* Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4. (Contributed by NM, 25-Mar-2004.)
((𝑤𝑥𝑧𝑤) → ((𝑧 (𝑥 ∖ {𝑧})) ∩ (𝑤 (𝑥 ∖ {𝑤}))) = ∅)
 
Theoremkmlem6 9842* Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 4 => 1. (Contributed by NM, 26-Mar-2004.)
((∀𝑧𝑥 𝑧 ≠ ∅ ∧ ∀𝑧𝑥𝑤𝑥 (𝜑𝐴 = ∅)) → ∀𝑧𝑥𝑣𝑧𝑤𝑥 (𝜑 → ¬ 𝑣𝐴))
 
Theoremkmlem7 9843* Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 4 => 1. (Contributed by NM, 26-Mar-2004.)
((∀𝑧𝑥 𝑧 ≠ ∅ ∧ ∀𝑧𝑥𝑤𝑥 (𝑧𝑤 → (𝑧𝑤) = ∅)) → ¬ ∃𝑧𝑥𝑣𝑧𝑤𝑥 (𝑧𝑤𝑣 ∈ (𝑧𝑤)))
 
Theoremkmlem8 9844* Lemma for 5-quantifier AC of Kurt Maes, Th. 4 1 <=> 4. (Contributed by NM, 4-Apr-2004.)
((¬ ∃𝑧𝑢𝑤𝑧 𝜓 → ∃𝑦𝑧𝑢 (𝑧 ≠ ∅ → ∃!𝑤 𝑤 ∈ (𝑧𝑦))) ↔ (∃𝑧𝑢𝑤𝑧 𝜓 ∨ ∃𝑦𝑦𝑢 ∧ ∀𝑧𝑢 ∃!𝑤 𝑤 ∈ (𝑧𝑦))))
 
Theoremkmlem9 9845* Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4. (Contributed by NM, 25-Mar-2004.)
𝐴 = {𝑢 ∣ ∃𝑡𝑥 𝑢 = (𝑡 (𝑥 ∖ {𝑡}))}       𝑧𝐴𝑤𝐴 (𝑧𝑤 → (𝑧𝑤) = ∅)
 
Theoremkmlem10 9846* Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4. (Contributed by NM, 25-Mar-2004.)
𝐴 = {𝑢 ∣ ∃𝑡𝑥 𝑢 = (𝑡 (𝑥 ∖ {𝑡}))}       (∀(∀𝑧𝑤 (𝑧𝑤 → (𝑧𝑤) = ∅) → ∃𝑦𝑧 𝜑) → ∃𝑦𝑧𝐴 𝜑)
 
Theoremkmlem11 9847* Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4. (Contributed by NM, 26-Mar-2004.)
𝐴 = {𝑢 ∣ ∃𝑡𝑥 𝑢 = (𝑡 (𝑥 ∖ {𝑡}))}       (𝑧𝑥 → (𝑧 𝐴) = (𝑧 (𝑥 ∖ {𝑧})))
 
Theoremkmlem12 9848* Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4. (Contributed by NM, 27-Mar-2004.)
𝐴 = {𝑢 ∣ ∃𝑡𝑥 𝑢 = (𝑡 (𝑥 ∖ {𝑡}))}       (∀𝑧𝑥 (𝑧 (𝑥 ∖ {𝑧})) ≠ ∅ → (∀𝑧𝐴 (𝑧 ≠ ∅ → ∃!𝑣 𝑣 ∈ (𝑧𝑦)) → ∀𝑧𝑥 (𝑧 ≠ ∅ → ∃!𝑣 𝑣 ∈ (𝑧 ∩ (𝑦 𝐴)))))
 
Theoremkmlem13 9849* Lemma for 5-quantifier AC of Kurt Maes, Th. 4 1 <=> 4. (Contributed by NM, 5-Apr-2004.)
𝐴 = {𝑢 ∣ ∃𝑡𝑥 𝑢 = (𝑡 (𝑥 ∖ {𝑡}))}       (∀𝑥((∀𝑧𝑥 𝑧 ≠ ∅ ∧ ∀𝑧𝑥𝑤𝑥 (𝑧𝑤 → (𝑧𝑤) = ∅)) → ∃𝑦𝑧𝑥 ∃!𝑣 𝑣 ∈ (𝑧𝑦)) ↔ ∀𝑥(¬ ∃𝑧𝑥𝑣𝑧𝑤𝑥 (𝑧𝑤𝑣 ∈ (𝑧𝑤)) → ∃𝑦𝑧𝑥 (𝑧 ≠ ∅ → ∃!𝑣 𝑣 ∈ (𝑧𝑦))))
 
Theoremkmlem14 9850* Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 5 <=> 4. (Contributed by NM, 4-Apr-2004.)
(𝜑 ↔ (𝑧𝑦 → ((𝑣𝑥𝑦𝑣) ∧ 𝑧𝑣)))    &   (𝜓 ↔ (𝑧𝑥 → ((𝑣𝑧𝑣𝑦) ∧ ((𝑢𝑧𝑢𝑦) → 𝑢 = 𝑣))))    &   (𝜒 ↔ ∀𝑧𝑥 ∃!𝑣 𝑣 ∈ (𝑧𝑦))       (∃𝑧𝑥𝑣𝑧𝑤𝑥 (𝑧𝑤𝑣 ∈ (𝑧𝑤)) ↔ ∃𝑦𝑧𝑣𝑢(𝑦𝑥𝜑))
 
Theoremkmlem15 9851* Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 5 <=> 4. (Contributed by NM, 4-Apr-2004.)
(𝜑 ↔ (𝑧𝑦 → ((𝑣𝑥𝑦𝑣) ∧ 𝑧𝑣)))    &   (𝜓 ↔ (𝑧𝑥 → ((𝑣𝑧𝑣𝑦) ∧ ((𝑢𝑧𝑢𝑦) → 𝑢 = 𝑣))))    &   (𝜒 ↔ ∀𝑧𝑥 ∃!𝑣 𝑣 ∈ (𝑧𝑦))       ((¬ 𝑦𝑥𝜒) ↔ ∀𝑧𝑣𝑢𝑦𝑥𝜓))
 
Theoremkmlem16 9852* Lemma for 5-quantifier AC of Kurt Maes, Th. 4 5 <=> 4. (Contributed by NM, 4-Apr-2004.)
(𝜑 ↔ (𝑧𝑦 → ((𝑣𝑥𝑦𝑣) ∧ 𝑧𝑣)))    &   (𝜓 ↔ (𝑧𝑥 → ((𝑣𝑧𝑣𝑦) ∧ ((𝑢𝑧𝑢𝑦) → 𝑢 = 𝑣))))    &   (𝜒 ↔ ∀𝑧𝑥 ∃!𝑣 𝑣 ∈ (𝑧𝑦))       ((∃𝑧𝑥𝑣𝑧𝑤𝑥 (𝑧𝑤𝑣 ∈ (𝑧𝑤)) ∨ ∃𝑦𝑦𝑥𝜒)) ↔ ∃𝑦𝑧𝑣𝑢((𝑦𝑥𝜑) ∨ (¬ 𝑦𝑥𝜓)))
 
Theoremdfackm 9853* Equivalence of the Axiom of Choice and Maes' AC ackm 10152. The proof consists of lemmas kmlem1 9837 through kmlem16 9852 and this final theorem. AC is not used for the proof. Note: bypassing the first step (i.e., replacing dfac5 9815 with biid 260) 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 ↔ ∀𝑥𝑦𝑧𝑣𝑢((𝑦𝑥 ∧ (𝑧𝑦 → ((𝑣𝑥 ∧ ¬ 𝑦 = 𝑣) ∧ 𝑧𝑣))) ∨ (¬ 𝑦𝑥 ∧ (𝑧𝑥 → ((𝑣𝑧𝑣𝑦) ∧ ((𝑢𝑧𝑢𝑦) → 𝑢 = 𝑣))))))
 
2.6.13  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 9590 () for cardinal addition, Cartesian product df-xp 5586 (×) for cardinal multiplication, and set exponentiation df-map 8575 (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 10238, carddom 10241, and cardsdom 10242. The advantage of Mendelson's approach is that we can directly use many equinumerosity theorems that we already have available.

 
Theoremundjudom 9854 Cardinal addition dominates union. (Contributed by NM, 28-Sep-2004.) (Revised by Jim Kingdon, 15-Aug-2023.)
((𝐴𝑉𝐵𝑊) → (𝐴𝐵) ≼ (𝐴𝐵))
 
Theoremendjudisj 9855 Equinumerosity of a disjoint union and a union of two disjoint sets. (Contributed by NM, 5-Apr-2007.)
((𝐴𝑉𝐵𝑊 ∧ (𝐴𝐵) = ∅) → (𝐴𝐵) ≈ (𝐴𝐵))
 
Theoremdjuen 9856 Disjoint unions of equinumerous sets are equinumerous. (Contributed by NM, 28-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
((𝐴𝐵𝐶𝐷) → (𝐴𝐶) ≈ (𝐵𝐷))
 
Theoremdjuenun 9857 Disjoint union is equinumerous to union for disjoint sets. (Contributed by Mario Carneiro, 29-Apr-2015.) (Revised by Jim Kingdon, 19-Aug-2023.)
((𝐴𝐵𝐶𝐷 ∧ (𝐵𝐷) = ∅) → (𝐴𝐶) ≈ (𝐵𝐷))
 
Theoremdju1en 9858 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.)
((𝐴𝑉 ∧ ¬ 𝐴𝐴) → (𝐴 ⊔ 1o) ≈ suc 𝐴)
 
Theoremdju1dif 9859 Adding and subtracting one gives back the original cardinality. Similar to pncan 11157 for cardinalities. (Contributed by Mario Carneiro, 18-May-2015.) (Revised by Jim Kingdon, 20-Aug-2023.)
((𝐴𝑉𝐵 ∈ (𝐴 ⊔ 1o)) → ((𝐴 ⊔ 1o) ∖ {𝐵}) ≈ 𝐴)
 
Theoremdju1p1e2 9860 1+1=2 for cardinal number addition, derived from pm54.43 9690 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 9584), 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 9861 for a shorter proof that doesn't use pm54.43 9690. (Contributed by NM, 5-Apr-2007.) (Proof modification is discouraged.)
(1o ⊔ 1o) ≈ 2o
 
Theoremdju1p1e2ALT 9861 Alternate proof of dju1p1e2 9860. (Contributed by Mario Carneiro, 29-Apr-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
(1o ⊔ 1o) ≈ 2o
 
Theoremdju0en 9862 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.)
(𝐴𝑉 → (𝐴 ⊔ ∅) ≈ 𝐴)
 
Theoremxp2dju 9863 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) = (𝐴𝐴)
 
Theoremdjucomen 9864 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.)
((𝐴𝑉𝐵𝑊) → (𝐴𝐵) ≈ (𝐵𝐴))
 
Theoremdjuassen 9865 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.)
((𝐴𝑉𝐵𝑊𝐶𝑋) → ((𝐴𝐵) ⊔ 𝐶) ≈ (𝐴 ⊔ (𝐵𝐶)))
 
Theoremxpdjuen 9866 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.)
((𝐴𝑉𝐵𝑊𝐶𝑋) → (𝐴 × (𝐵𝐶)) ≈ ((𝐴 × 𝐵) ⊔ (𝐴 × 𝐶)))
 
Theoremmapdjuen 9867 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.)
((𝐴𝑉𝐵𝑊𝐶𝑋) → (𝐴m (𝐵𝐶)) ≈ ((𝐴m 𝐵) × (𝐴m 𝐶)))
 
Theorempwdjuen 9868 Sum of exponents law for cardinal arithmetic. (Contributed by Mario Carneiro, 15-May-2015.)
((𝐴𝑉𝐵𝑊) → 𝒫 (𝐴𝐵) ≈ (𝒫 𝐴 × 𝒫 𝐵))
 
Theoremdjudom1 9869 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.)
((𝐴𝐵𝐶𝑉) → (𝐴𝐶) ≼ (𝐵𝐶))
 
Theoremdjudom2 9870 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.)
((𝐴𝐵𝐶𝑉) → (𝐶𝐴) ≼ (𝐶𝐵))
 
Theoremdjudoml 9871 A set is dominated by its disjoint union with another. (Contributed by NM, 28-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
((𝐴𝑉𝐵𝑊) → 𝐴 ≼ (𝐴𝐵))
 
Theoremdjuxpdom 9872 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.)
((1o𝐴 ∧ 1o𝐵) → (𝐴𝐵) ≼ (𝐴 × 𝐵))
 
Theoremdjufi 9873 The disjoint union of two finite sets is finite. (Contributed by NM, 22-Oct-2004.)
((𝐴 ≺ ω ∧ 𝐵 ≺ ω) → (𝐴𝐵) ≺ ω)
 
Theoremcdainflem 9874 Any partition of omega into two pieces (which may be disjoint) contains an infinite subset. (Contributed by Mario Carneiro, 11-Feb-2013.)
((𝐴𝐵) ≈ ω → (𝐴 ≈ ω ∨ 𝐵 ≈ ω))
 
Theoremdjuinf 9875 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 ≼ (𝐴𝐴))
 
Theoreminfdju1 9876 An infinite set is equinumerous to itself added with one. (Contributed by Mario Carneiro, 15-May-2015.)
(ω ≼ 𝐴 → (𝐴 ⊔ 1o) ≈ 𝐴)
 
Theorempwdju1 9877 The sum of a powerset with itself is equipotent to the successor powerset. (Contributed by Mario Carneiro, 15-May-2015.)
(𝐴𝑉 → (𝒫 𝐴 ⊔ 𝒫 𝐴) ≈ 𝒫 (𝐴 ⊔ 1o))
 
Theorempwdjuidm 9878 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)
 
Theoremdjulepw 9879 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.)
(((𝐴𝐴) ≈ 𝐴𝐵 ≼ 𝒫 𝐴) → (𝐴𝐵) ≼ 𝒫 𝐴)
 
Theoremonadju 9880 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) ≈ (𝐴𝐵))
 
Theoremcardadju 9881 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) → (𝐴𝐵) ≈ ((card‘𝐴) +o (card‘𝐵)))
 
Theoremdjunum 9882 The disjoint union of two numerable sets is numerable. (Contributed by Mario Carneiro, 29-Apr-2015.)
((𝐴 ∈ dom card ∧ 𝐵 ∈ dom card) → (𝐴𝐵) ∈ dom card)
 
Theoremunnum 9883 The union of two numerable sets is numerable. (Contributed by Mario Carneiro, 29-Apr-2015.)
((𝐴 ∈ dom card ∧ 𝐵 ∈ dom card) → (𝐴𝐵) ∈ dom card)
 
Theoremnnadju 9884 The cardinal and ordinal sums of finite ordinals are equal. For a shorter proof using ax-rep 5205, see nnadjuALT 9885. (Contributed by Paul Chapman, 11-Apr-2009.) (Revised by Mario Carneiro, 6-Feb-2013.) Avoid ax-rep 5205. (Revised by BTernaryTau, 2-Jul-2024.)
((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (card‘(𝐴𝐵)) = (𝐴 +o 𝐵))
 
TheoremnnadjuALT 9885 Shorter proof of nnadju 9884 using ax-rep 5205. (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‘(𝐴𝐵)) = (𝐴 +o 𝐵))
 
Theoremficardadju 9886 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) → (𝐴𝐵) ≈ ((card‘𝐴) +o (card‘𝐵)))
 
Theoremficardun 9887 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 5205. (Revised by BTernaryTau, 3-Jul-2024.)
((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin ∧ (𝐴𝐵) = ∅) → (card‘(𝐴𝐵)) = ((card‘𝐴) +o (card‘𝐵)))
 
TheoremficardunOLD 9888 Obsolete version of ficardun 9887 as of 3-Jul-2024. (Contributed by Paul Chapman, 5-Jun-2009.) (Proof shortened by Mario Carneiro, 28-Apr-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin ∧ (𝐴𝐵) = ∅) → (card‘(𝐴𝐵)) = ((card‘𝐴) +o (card‘𝐵)))
 
Theoremficardun2 9889 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 5205. (Revised by BTernaryTau, 3-Jul-2024.)
((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) → (card‘(𝐴𝐵)) ⊆ ((card‘𝐴) +o (card‘𝐵)))
 
Theoremficardun2OLD 9890 Obsolete version of ficardun2 9889 as of 3-Jul-2024. (Contributed by Mario Carneiro, 5-Feb-2013.) (Proof modification is discouraged.) (New usage is discouraged.)
((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) → (card‘(𝐴𝐵)) ⊆ ((card‘𝐴) +o (card‘𝐵)))
 
Theorempwsdompw 9891* Lemma for domtriom 10130. This is the equinumerosity version of the algebraic identity Σ𝑘𝑛(2↑𝑘) = (2↑𝑛) − 1. (Contributed by Mario Carneiro, 7-Feb-2013.)
((𝑛 ∈ ω ∧ ∀𝑘 ∈ suc 𝑛(𝐵𝑘) ≈ 𝒫 𝑘) → 𝑘𝑛 (𝐵𝑘) ≺ (𝐵𝑛))
 
Theoremunctb 9892 The union of two countable sets is countable. (Contributed by FL, 25-Aug-2006.) (Proof shortened by Mario Carneiro, 30-Apr-2015.)
((𝐴 ≼ ω ∧ 𝐵 ≼ ω) → (𝐴𝐵) ≼ ω)
 
Theoreminfdjuabs 9893 Absorption law for addition to an infinite cardinal. (Contributed by NM, 30-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
((𝐴 ∈ dom card ∧ ω ≼ 𝐴𝐵𝐴) → (𝐴𝐵) ≈ 𝐴)
 
Theoreminfunabs 9894 An infinite set is equinumerous to its union with a smaller one. (Contributed by NM, 28-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
((𝐴 ∈ dom card ∧ ω ≼ 𝐴𝐵𝐴) → (𝐴𝐵) ≈ 𝐴)
 
Theoreminfdju 9895 The sum of two cardinal numbers is their maximum, if one of them is infinite. Proposition 10.41 of [TakeutiZaring] p. 95. (Contributed by NM, 28-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
((𝐴 ∈ dom card ∧ 𝐵 ∈ dom card ∧ ω ≼ 𝐴) → (𝐴𝐵) ≈ (𝐴𝐵))
 
Theoreminfdif 9896 The cardinality of an infinite set does not change after subtracting a strictly smaller one. Example in [Enderton] p. 164. (Contributed by NM, 22-Oct-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
((𝐴 ∈ dom card ∧ ω ≼ 𝐴𝐵𝐴) → (𝐴𝐵) ≈ 𝐴)
 
Theoreminfdif2 9897 Cardinality ordering for an infinite class difference. (Contributed by NM, 24-Mar-2007.) (Revised by Mario Carneiro, 29-Apr-2015.)
((𝐴 ∈ dom card ∧ 𝐵 ∈ dom card ∧ ω ≼ 𝐴) → ((𝐴𝐵) ≼ 𝐵𝐴𝐵))
 
Theoreminfxpdom 9898 Dominance law for multiplication with an infinite cardinal. (Contributed by NM, 26-Mar-2006.) (Revised by Mario Carneiro, 29-Apr-2015.)
((𝐴 ∈ dom card ∧ ω ≼ 𝐴𝐵𝐴) → (𝐴 × 𝐵) ≼ 𝐴)
 
Theoreminfxpabs 9899 Absorption law for multiplication with an infinite cardinal. (Contributed by NM, 30-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
(((𝐴 ∈ dom card ∧ ω ≼ 𝐴) ∧ (𝐵 ≠ ∅ ∧ 𝐵𝐴)) → (𝐴 × 𝐵) ≈ 𝐴)
 
Theoreminfunsdom1 9900 The union of two sets that are strictly dominated by the infinite set 𝑋 is also dominated by 𝑋. This version of infunsdom 9901 assumes additionally that 𝐴 is the smaller of the two. (Contributed by Mario Carneiro, 14-Dec-2013.) (Revised by Mario Carneiro, 3-May-2015.)
(((𝑋 ∈ dom card ∧ ω ≼ 𝑋) ∧ (𝐴𝐵𝐵𝑋)) → (𝐴𝐵) ≺ 𝑋)
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330 32901-33000 331 33001-33100 332 33101-33200 333 33201-33300 334 33301-33400 335 33401-33500 336 33501-33600 337 33601-33700 338 33701-33800 339 33801-33900 340 33901-34000 341 34001-34100 342 34101-34200 343 34201-34300 344 34301-34400 345 34401-34500 346 34501-34600 347 34601-34700 348 34701-34800 349 34801-34900 350 34901-35000 351 35001-35100 352 35101-35200 353 35201-35300 354 35301-35400 355 35401-35500 356 35501-35600 357 35601-35700 358 35701-35800 359 35801-35900 360 35901-36000 361 36001-36100 362 36101-36200 363 36201-36300 364 36301-36400 365 36401-36500 366 36501-36600 367 36601-36700 368 36701-36800 369 36801-36900 370 36901-37000 371 37001-37100 372 37101-37200 373 37201-37300 374 37301-37400 375 37401-37500 376 37501-37600 377 37601-37700 378 37701-37800 379 37801-37900 380 37901-38000 381 38001-38100 382 38101-38200 383 38201-38300 384 38301-38400 385 38401-38500 386 38501-38600 387 38601-38700 388 38701-38800 389 38801-38900 390 38901-39000 391 39001-39100 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