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| Type | Label | Description |
|---|---|---|
| Statement | ||
| Theorem | alephdom2 10001 | 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 10002 | The argument of the aleph function is less than or equal to its value. Exercise 2 of [TakeutiZaring] p. 91. (Later, in alephfp2 10023, 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 10003* | 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 10004* | 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 10005* | 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 10006 | 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 10007 | 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 10008 | 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 10009* | 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 10010 | 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 10011* | 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 10012 | 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 10013 | 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 10014 | 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 10015 | 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 10016 | The aleph function is strictly monotone. (Contributed by Mario Carneiro, 15-Mar-2013.) |
| ⊢ Smo ℵ | ||
| Theorem | alephf1ALT 10017 | Alternate proof of alephf1 9999. (Contributed by Mario Carneiro, 15-Mar-2013.) (Proof modification is discouraged.) (New usage is discouraged.) |
| ⊢ ℵ:On–1-1→On | ||
| Theorem | alephfplem1 10018 | Lemma for alephfp 10022. (Contributed by NM, 6-Nov-2004.) |
| ⊢ 𝐻 = (rec(ℵ, ω) ↾ ω) ⇒ ⊢ (𝐻‘∅) ∈ ran ℵ | ||
| Theorem | alephfplem2 10019* | Lemma for alephfp 10022. (Contributed by NM, 6-Nov-2004.) |
| ⊢ 𝐻 = (rec(ℵ, ω) ↾ ω) ⇒ ⊢ (𝑤 ∈ ω → (𝐻‘suc 𝑤) = (ℵ‘(𝐻‘𝑤))) | ||
| Theorem | alephfplem3 10020* | Lemma for alephfp 10022. (Contributed by NM, 6-Nov-2004.) |
| ⊢ 𝐻 = (rec(ℵ, ω) ↾ ω) ⇒ ⊢ (𝑣 ∈ ω → (𝐻‘𝑣) ∈ ran ℵ) | ||
| Theorem | alephfplem4 10021 | Lemma for alephfp 10022. (Contributed by NM, 5-Nov-2004.) |
| ⊢ 𝐻 = (rec(ℵ, ω) ↾ ω) ⇒ ⊢ ∪ (𝐻 “ ω) ∈ ran ℵ | ||
| Theorem | alephfp 10022 | 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 10023 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 10023 | The aleph function has at least one fixed point. Proposition 11.18 of [TakeutiZaring] p. 104. See alephfp 10022 for an actual example of a fixed point. Compare the inequality alephle 10002 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 10024* | 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 10025 | The power set of an aleph is not strictly dominated by the successor aleph. (The Generalized Continuum Hypothesis says they are equinumerous, see gch3 10591 or gchaleph2 10587.) The transposed form alephsucpw 10485 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 10026 | 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 10027* | 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 10028 | 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 10029 | Wff for an abbreviation of the axiom of choice. |
| wff CHOICE | ||
| Definition | df-ac 10030* |
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 10373 as our definition, because the equivalence to more standard forms (dfac2 10046) 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 10373 itself as dfac0 10048. (Contributed by Mario Carneiro, 22-Feb-2015.) |
| ⊢ (CHOICE ↔ ∀𝑥∃𝑓(𝑓 ⊆ 𝑥 ∧ 𝑓 Fn dom 𝑥)) | ||
| Theorem | aceq1 10031* | Equivalence of two versions of the Axiom of Choice ax-ac 10373. 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 10032* | 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 10373. (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 10033* | 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 10034* | Lemma for dfac3 10035. (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 10035* | 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 10036* | 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 10037* | Lemma for dfac5 10043. (Contributed by NM, 12-Apr-2004.) |
| ⊢ (∃!𝑣 𝑣 ∈ (({𝑤} × 𝑤) ∩ 𝑦) ↔ ∃!𝑔(𝑔 ∈ 𝑤 ∧ 〈𝑤, 𝑔〉 ∈ 𝑦)) | ||
| Theorem | dfac5lem2 10038* | Lemma for dfac5 10043. (Contributed by NM, 12-Apr-2004.) |
| ⊢ 𝐴 = {𝑢 ∣ (𝑢 ≠ ∅ ∧ ∃𝑡 ∈ ℎ 𝑢 = ({𝑡} × 𝑡))} ⇒ ⊢ (〈𝑤, 𝑔〉 ∈ ∪ 𝐴 ↔ (𝑤 ∈ ℎ ∧ 𝑔 ∈ 𝑤)) | ||
| Theorem | dfac5lem3 10039* | Lemma for dfac5 10043. (Contributed by NM, 12-Apr-2004.) |
| ⊢ 𝐴 = {𝑢 ∣ (𝑢 ≠ ∅ ∧ ∃𝑡 ∈ ℎ 𝑢 = ({𝑡} × 𝑡))} ⇒ ⊢ (({𝑤} × 𝑤) ∈ 𝐴 ↔ (𝑤 ≠ ∅ ∧ 𝑤 ∈ ℎ)) | ||
| Theorem | dfac5lem4 10040* | Lemma for dfac5 10043. (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 10041* | Lemma for dfac5 10043. (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 10042* | Obsolete version of dfac5lem4 10040 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 10043* | 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 10044* | Our Axiom of Choice (in the form of ac3 10376) implies the Axiom of Choice (first form) of [Enderton] p. 49. The proof uses neither AC nor the Axiom of Regularity. See dfac2b 10045 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 10045* | Axiom of Choice (first form) of [Enderton] p. 49 implies our Axiom of Choice (in the form of ac3 10376). The proof does not make use of AC. Note that the Axiom of Regularity is used by the proof. Specifically, elneq 9509 and preleq 9529 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 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 | dfac2 10046* | Axiom of Choice (first form) of [Enderton] p. 49 corresponds to our Axiom of Choice (in the form of ac3 10376). The proof does not make use of AC, but the Axiom of Regularity is used (by applying dfac2b 10045). (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 10047* | Equivalence of the Axiom of Choice (first form) of [Enderton] p. 49 and our Axiom of Choice (in the form of ac2 10375). 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 10048* | 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 10373. (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 10049* | Equivalence of two versions of the Axiom of Choice ax-ac 10373. 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 10050* | A proof of the equivalency of the well-ordering theorem weth 10409 and the axiom of choice ac7 10387. (Contributed by Mario Carneiro, 5-Jan-2013.) |
| ⊢ (CHOICE ↔ ∀𝑥∃𝑟 𝑟 We 𝑥) | ||
| Theorem | dfac9 10051* | Equivalence of the axiom of choice with a statement related to ac9 10397; 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 10052 | Axiom of Choice equivalent: the cardinality function measures every set. (Contributed by Mario Carneiro, 6-May-2015.) |
| ⊢ (CHOICE ↔ dom card = V) | ||
| Theorem | dfac10c 10053* | 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 10054 | Axiom of Choice equivalent: every set is equinumerous to an ordinal (quantifier-free short cryptic version alluded to in df-ac 10030). (Contributed by Stefan O'Rear, 17-Jan-2015.) |
| ⊢ (CHOICE ↔ ( ≈ “ On) = V) | ||
| Theorem | acacni 10055 | 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 10056 | A choice equivalent: every set has choice sets of every length. (Contributed by Mario Carneiro, 31-Aug-2015.) |
| ⊢ (CHOICE ↔ ∀𝑥AC 𝑥 = V) | ||
| Theorem | dfac13 10057 | 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 10058* | Lemma for dfac12 10064. (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 10059* | Lemma for dfac12 10064. (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 10060* | Lemma for dfac12 10064. (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 10061 | The axiom of choice holds iff every ordinal has a well-orderable powerset. This version of dfac12 10064 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 10062* | Equivalence of dfac12 10064 and dfac12a 10063, 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 10063 | 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 10064 | 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 10065* | 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 10066* | 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 10067* | 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 10068* | 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 10069* | 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 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 (𝜑 → 𝐴 = ∅)) → ∀𝑧 ∈ 𝑥 ∃𝑣 ∈ 𝑧 ∀𝑤 ∈ 𝑥 (𝜑 → ¬ 𝑣 ∈ 𝐴)) | ||
| Theorem | kmlem7 10071* | 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 10072* | 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 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}))} ⇒ ⊢ ∀𝑧 ∈ 𝐴 ∀𝑤 ∈ 𝐴 (𝑧 ≠ 𝑤 → (𝑧 ∩ 𝑤) = ∅) | ||
| Theorem | kmlem10 10074* | 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 10075* | 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 10076* | 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 10077* | 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 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)) ⇒ ⊢ (∃𝑧 ∈ 𝑥 ∀𝑣 ∈ 𝑧 ∃𝑤 ∈ 𝑥 (𝑧 ≠ 𝑤 ∧ 𝑣 ∈ (𝑧 ∩ 𝑤)) ↔ ∃𝑦∀𝑧∃𝑣∀𝑢(𝑦 ∈ 𝑥 ∧ 𝜑)) | ||
| Theorem | kmlem15 10079* | 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 10080* | 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 10081* | Equivalence of the Axiom of Choice and Maes' AC ackm 10379. The proof consists of lemmas kmlem1 10065 through kmlem16 10080 and this final theorem. AC is not used for the proof. Note: bypassing the first step (i.e., replacing dfac5 10043 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 9817 (⊔) for cardinal addition, Cartesian product df-xp 5631 (×) for cardinal multiplication, and set exponentiation df-map 8769 (↑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 10465, carddom 10468, and cardsdom 10469. The advantage of Mendelson's approach is that we can directly use many equinumerosity theorems that we already have available. | ||
| Theorem | undjudom 10082 | 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 10083 | 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 10084 | 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 10085 | 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 10086 | 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 10087 | Adding and subtracting one gives back the original cardinality. Similar to pncan 11390 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 10088 | 1+1=2 for cardinal number addition, derived from pm54.43 9917 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 9811), 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 10089 for a shorter proof that doesn't use pm54.43 9917. (Contributed by NM, 5-Apr-2007.) (Proof modification is discouraged.) |
| ⊢ (1o ⊔ 1o) ≈ 2o | ||
| Theorem | dju1p1e2ALT 10089 | Alternate proof of dju1p1e2 10088. (Contributed by Mario Carneiro, 29-Apr-2015.) (Proof modification is discouraged.) (New usage is discouraged.) |
| ⊢ (1o ⊔ 1o) ≈ 2o | ||
| Theorem | dju0en 10090 | 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 10091 | 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 10092 | 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 10093 | 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 10094 | 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 10095 | 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 10096 | 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 10097 | 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 10098 | 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 10099 | 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 10100 | 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)) | ||
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