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| Type | Label | Description |
|---|---|---|
| Statement | ||
| Theorem | alephle 10001 | The argument of the aleph function is less than or equal to its value. Exercise 2 of [TakeutiZaring] p. 91. (Later, in alephfp2 10022, 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 10002* | 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 10003* | 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 10004* | 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 10005 | 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 10006 | 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 10007 | 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 10008* | 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 10009 | 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 10010* | 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 10011 | 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 10012 | 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 10013 | 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 10014 | 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 10015 | The aleph function is strictly monotone. (Contributed by Mario Carneiro, 15-Mar-2013.) |
| ⊢ Smo ℵ | ||
| Theorem | alephf1ALT 10016 | Alternate proof of alephf1 9998. (Contributed by Mario Carneiro, 15-Mar-2013.) (Proof modification is discouraged.) (New usage is discouraged.) |
| ⊢ ℵ:On–1-1→On | ||
| Theorem | alephfplem1 10017 | Lemma for alephfp 10021. (Contributed by NM, 6-Nov-2004.) |
| ⊢ 𝐻 = (rec(ℵ, ω) ↾ ω) ⇒ ⊢ (𝐻‘∅) ∈ ran ℵ | ||
| Theorem | alephfplem2 10018* | Lemma for alephfp 10021. (Contributed by NM, 6-Nov-2004.) |
| ⊢ 𝐻 = (rec(ℵ, ω) ↾ ω) ⇒ ⊢ (𝑤 ∈ ω → (𝐻‘suc 𝑤) = (ℵ‘(𝐻‘𝑤))) | ||
| Theorem | alephfplem3 10019* | Lemma for alephfp 10021. (Contributed by NM, 6-Nov-2004.) |
| ⊢ 𝐻 = (rec(ℵ, ω) ↾ ω) ⇒ ⊢ (𝑣 ∈ ω → (𝐻‘𝑣) ∈ ran ℵ) | ||
| Theorem | alephfplem4 10020 | Lemma for alephfp 10021. (Contributed by NM, 5-Nov-2004.) |
| ⊢ 𝐻 = (rec(ℵ, ω) ↾ ω) ⇒ ⊢ ∪ (𝐻 “ ω) ∈ ran ℵ | ||
| Theorem | alephfp 10021 | 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 10022 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 10022 | The aleph function has at least one fixed point. Proposition 11.18 of [TakeutiZaring] p. 104. See alephfp 10021 for an actual example of a fixed point. Compare the inequality alephle 10001 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 10023* | 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 10024 | The power set of an aleph is not strictly dominated by the successor aleph. (The Generalized Continuum Hypothesis says they are equinumerous, see gch3 10589 or gchaleph2 10585.) The transposed form alephsucpw 10483 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 10025 | 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 10026* | 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 10027 | 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) | ||
| Syntax | wac 10028 | Wff for an abbreviation of the axiom of choice. |
| wff CHOICE | ||
| Definition | df-ac 10029* |
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 10372 as our definition, because the equivalence to more standard forms (dfac2 10045) 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 10372 itself as dfac0 10047. (Contributed by Mario Carneiro, 22-Feb-2015.) |
| ⊢ (CHOICE ↔ ∀𝑥∃𝑓(𝑓 ⊆ 𝑥 ∧ 𝑓 Fn dom 𝑥)) | ||
| Theorem | aceq1 10030* | Equivalence of two versions of the Axiom of Choice ax-ac 10372. 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 10031* | 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 10372. (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 10032* | 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 10033* | Lemma for dfac3 10034. (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 10034* | 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 10035* | 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 10036* | Lemma for dfac5 10042. (Contributed by NM, 12-Apr-2004.) |
| ⊢ (∃!𝑣 𝑣 ∈ (({𝑤} × 𝑤) ∩ 𝑦) ↔ ∃!𝑔(𝑔 ∈ 𝑤 ∧ 〈𝑤, 𝑔〉 ∈ 𝑦)) | ||
| Theorem | dfac5lem2 10037* | Lemma for dfac5 10042. (Contributed by NM, 12-Apr-2004.) |
| ⊢ 𝐴 = {𝑢 ∣ (𝑢 ≠ ∅ ∧ ∃𝑡 ∈ ℎ 𝑢 = ({𝑡} × 𝑡))} ⇒ ⊢ (〈𝑤, 𝑔〉 ∈ ∪ 𝐴 ↔ (𝑤 ∈ ℎ ∧ 𝑔 ∈ 𝑤)) | ||
| Theorem | dfac5lem3 10038* | Lemma for dfac5 10042. (Contributed by NM, 12-Apr-2004.) |
| ⊢ 𝐴 = {𝑢 ∣ (𝑢 ≠ ∅ ∧ ∃𝑡 ∈ ℎ 𝑢 = ({𝑡} × 𝑡))} ⇒ ⊢ (({𝑤} × 𝑤) ∈ 𝐴 ↔ (𝑤 ≠ ∅ ∧ 𝑤 ∈ ℎ)) | ||
| Theorem | dfac5lem4 10039* | Lemma for dfac5 10042. (Contributed by NM, 11-Apr-2004.) Avoid ax-11 2158. (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 10040* | Lemma for dfac5 10042. (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 10041* | Obsolete version of dfac5lem4 10039 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 10042* | 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 10043* | Our Axiom of Choice (in the form of ac3 10375) implies the Axiom of Choice (first form) of [Enderton] p. 49. The proof uses neither AC nor the Axiom of Regularity. See dfac2b 10044 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 10044* | Axiom of Choice (first form) of [Enderton] p. 49 implies our Axiom of Choice (in the form of ac3 10375). The proof does not make use of AC. Note that the Axiom of Regularity is used by the proof. Specifically, elneq 9511 and preleq 9531 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 10043.) (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 10045* | Axiom of Choice (first form) of [Enderton] p. 49 corresponds to our Axiom of Choice (in the form of ac3 10375). The proof does not make use of AC, but the Axiom of Regularity is used (by applying dfac2b 10044). (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 10046* | Equivalence of the Axiom of Choice (first form) of [Enderton] p. 49 and our Axiom of Choice (in the form of ac2 10374). 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 10047* | 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 10372. (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 10048* | Equivalence of two versions of the Axiom of Choice ax-ac 10372. 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 10049* | A proof of the equivalency of the well-ordering theorem weth 10408 and the axiom of choice ac7 10386. (Contributed by Mario Carneiro, 5-Jan-2013.) |
| ⊢ (CHOICE ↔ ∀𝑥∃𝑟 𝑟 We 𝑥) | ||
| Theorem | dfac9 10050* | Equivalence of the axiom of choice with a statement related to ac9 10396; 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 10051 | Axiom of Choice equivalent: the cardinality function measures every set. (Contributed by Mario Carneiro, 6-May-2015.) |
| ⊢ (CHOICE ↔ dom card = V) | ||
| Theorem | dfac10c 10052* | 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 10053 | Axiom of Choice equivalent: every set is equinumerous to an ordinal (quantifier-free short cryptic version alluded to in df-ac 10029). (Contributed by Stefan O'Rear, 17-Jan-2015.) |
| ⊢ (CHOICE ↔ ( ≈ “ On) = V) | ||
| Theorem | acacni 10054 | 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 10055 | A choice equivalent: every set has choice sets of every length. (Contributed by Mario Carneiro, 31-Aug-2015.) |
| ⊢ (CHOICE ↔ ∀𝑥AC 𝑥 = V) | ||
| Theorem | dfac13 10056 | 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 10057* | Lemma for dfac12 10063. (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 10058* | Lemma for dfac12 10063. (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 10059* | Lemma for dfac12 10063. (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 10060 | The axiom of choice holds iff every ordinal has a well-orderable powerset. This version of dfac12 10063 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 10061* | Equivalence of dfac12 10063 and dfac12a 10062, 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 10062 | 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 10063 | 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 10064* | 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 10065* | 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 10066* | 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 10067* | 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 10068* | 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 10069* | 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 10070* | 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 10071* | 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 10072* | 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 10073* | 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 10074* | 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 10075* | 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 10076* | 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 10077* | 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 10078* | 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 10079* | 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 10080* | Equivalence of the Axiom of Choice and Maes' AC ackm 10378. The proof consists of lemmas kmlem1 10064 through kmlem16 10079 and this final theorem. AC is not used for the proof. Note: bypassing the first step (i.e., replacing dfac5 10042 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)))))) | ||
For cardinal arithmetic, we follow [Mendelson] p. 258. Rather than defining operations restricted to cardinal numbers, we use disjoint union df-dju 9816 (⊔) for cardinal addition, Cartesian product df-xp 5629 (×) for cardinal multiplication, and set exponentiation df-map 8762 (↑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 10464, carddom 10467, and cardsdom 10468. The advantage of Mendelson's approach is that we can directly use many equinumerosity theorems that we already have available. | ||
| Theorem | undjudom 10081 | 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 10082 | 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 10083 | 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 10084 | 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 10085 | 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 10086 | Adding and subtracting one gives back the original cardinality. Similar to pncan 11387 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 10087 | 1+1=2 for cardinal number addition, derived from pm54.43 9916 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 9810), 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 10088 for a shorter proof that doesn't use pm54.43 9916. (Contributed by NM, 5-Apr-2007.) (Proof modification is discouraged.) |
| ⊢ (1o ⊔ 1o) ≈ 2o | ||
| Theorem | dju1p1e2ALT 10088 | Alternate proof of dju1p1e2 10087. (Contributed by Mario Carneiro, 29-Apr-2015.) (Proof modification is discouraged.) (New usage is discouraged.) |
| ⊢ (1o ⊔ 1o) ≈ 2o | ||
| Theorem | dju0en 10089 | 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 10090 | 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 10091 | 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 10092 | 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 10093 | 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 10094 | 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 10095 | 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 10096 | 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 10097 | 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 10098 | 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 10099 | 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 10100 | The disjoint union of two finite sets is finite. (Contributed by NM, 22-Oct-2004.) |
| ⊢ ((𝐴 ≺ ω ∧ 𝐵 ≺ ω) → (𝐴 ⊔ 𝐵) ≺ ω) | ||
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