P. 102 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 10101-10200   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	aceq0 10101*	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 10442. (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 10102*	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 10103*	Lemma for dfac3 10104. (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 10104*	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 10105*	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 10106*	Lemma for dfac5 10111. (Contributed by NM, 12-Apr-2004.)
⊢ (∃!𝑣 𝑣 ∈ (({𝑤} × 𝑤) ∩ 𝑦) ↔ ∃!𝑔(𝑔 ∈ 𝑤 ∧ ⟨𝑤, 𝑔⟩ ∈ 𝑦))
Theorem	dfac5lem2 10107*	Lemma for dfac5 10111. (Contributed by NM, 12-Apr-2004.)
⊢ 𝐴 = {𝑢 ∣ (𝑢 ≠ ∅ ∧ ∃𝑡 ∈ ℎ 𝑢 = ({𝑡} × 𝑡))}    ⇒   ⊢ (⟨𝑤, 𝑔⟩ ∈ ∪ 𝐴 ↔ (𝑤 ∈ ℎ ∧ 𝑔 ∈ 𝑤))
Theorem	dfac5lem3 10108*	Lemma for dfac5 10111. (Contributed by NM, 12-Apr-2004.)
⊢ 𝐴 = {𝑢 ∣ (𝑢 ≠ ∅ ∧ ∃𝑡 ∈ ℎ 𝑢 = ({𝑡} × 𝑡))}    ⇒   ⊢ (({𝑤} × 𝑤) ∈ 𝐴 ↔ (𝑤 ≠ ∅ ∧ 𝑤 ∈ ℎ))
Theorem	dfac5lem4 10109*	Lemma for dfac5 10111. (Contributed by NM, 11-Apr-2004.) Avoid ax-11 2190. (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 10110*	Lemma for dfac5 10111. (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	dfac5 10111*	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 10112*	Our Axiom of Choice (in the form of ac3 10445) implies the Axiom of Choice (first form) of [Enderton] p. 49. The proof uses neither AC nor the Axiom of Regularity. See dfac2b 10113 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 10113*	Axiom of Choice (first form) of [Enderton] p. 49 implies our Axiom of Choice (in the form of ac3 10445). The proof does not make use of AC. Note that the Axiom of Regularity is used by the proof. Specifically, elneq 9562 and preleq 9584 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 10112.) (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 10114*	Axiom of Choice (first form) of [Enderton] p. 49 corresponds to our Axiom of Choice (in the form of ac3 10445). The proof does not make use of AC, but the Axiom of Regularity is used (by applying dfac2b 10113). (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 10115*	Equivalence of the Axiom of Choice (first form) of [Enderton] p. 49 and our Axiom of Choice (in the form of ac2 10444). 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 10116*	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 10442. (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 10117*	Equivalence of two versions of the Axiom of Choice ax-ac 10442. 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 10118*	A proof of the equivalency of the well-ordering theorem weth 10478 and the axiom of choice ac7 10456. (Contributed by Mario Carneiro, 5-Jan-2013.)
⊢ (CHOICE ↔ ∀𝑥∃𝑟 𝑟 We 𝑥)
Theorem	dfac9 10119*	Equivalence of the axiom of choice with a statement related to ac9 10466; 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 10120	Axiom of Choice equivalent: the cardinality function measures every set. (Contributed by Mario Carneiro, 6-May-2015.)
⊢ (CHOICE ↔ dom card = V)
Theorem	dfac10c 10121*	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 10122	Axiom of Choice equivalent: every set is equinumerous to an ordinal (quantifier-free short cryptic version alluded to in df-ac 10099). (Contributed by Stefan O'Rear, 17-Jan-2015.)
⊢ (CHOICE ↔ ( ≈ “ On) = V)
Theorem	acacni 10123	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 10124	A choice equivalent: every set has choice sets of every length. (Contributed by Mario Carneiro, 31-Aug-2015.)
⊢ (CHOICE ↔ ∀𝑥AC 𝑥 = V)
Theorem	dfac13 10125	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 10126*	Lemma for dfac12 10132. (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 10127*	Lemma for dfac12 10132. (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 10128*	Lemma for dfac12 10132. (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 10129	The axiom of choice holds iff every ordinal has a well-orderable powerset. This version of dfac12 10132 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 10130*	Equivalence of dfac12 10132 and dfac12a 10131, 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 10131	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 10132	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 10133*	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 10134*	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 10135*	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 10136*	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 10137*	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 10138*	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 10139*	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 10140*	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 10141*	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 10142*	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 10143*	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 10144*	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 10145*	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 10146*	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 10147*	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 10148*	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 10149*	Equivalence of the Axiom of Choice and Maes' AC ackm 10448. The proof consists of lemmas kmlem1 10133 through kmlem16 10148 and this final theorem. AC is not used for the proof. Note: bypassing the first step (i.e., replacing dfac5 10111 with biid 264) 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 9886 (⊔) for cardinal addition, Cartesian product df-xp 5667 (×) for cardinal multiplication, and set exponentiation df-map 8825 (↑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 10534, carddom 10537, and cardsdom 10538. The advantage of Mendelson's approach is that we can directly use many equinumerosity theorems that we already have available.
Theorem	undjudom 10150	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 10151	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 10152	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 10153	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 10154	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 10155	Adding and subtracting one gives back the original cardinality. Similar to pncan 11462 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 10156	1+1=2 for cardinal number addition, derived from pm54.43 9986 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 9880), 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 10157 for a shorter proof that doesn't use pm54.43 9986. (Contributed by NM, 5-Apr-2007.) (Proof modification is discouraged.)
⊢ (1o ⊔ 1o) ≈ 2o
Theorem	dju1p1e2ALT 10157	Alternate proof of dju1p1e2 10156. (Contributed by Mario Carneiro, 29-Apr-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ (1o ⊔ 1o) ≈ 2o
Theorem	dju0en 10158	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 10159	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 10160	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 10161	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 10162	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 10163	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 10164	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 10165	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 10166	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 10167	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 10168	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 10169	The disjoint union of two finite sets is finite. (Contributed by NM, 22-Oct-2004.)
⊢ ((𝐴 ≺ ω ∧ 𝐵 ≺ ω) → (𝐴 ⊔ 𝐵) ≺ ω)
Theorem	cdainflem 10170	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 10171	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 10172	An infinite set is equinumerous to itself added with one. (Contributed by Mario Carneiro, 15-May-2015.)
⊢ (ω ≼ 𝐴 → (𝐴 ⊔ 1o) ≈ 𝐴)
Theorem	pwdju1 10173	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 10174	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 10175	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 10176	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 10177	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 10178	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 10179	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 10180	The cardinal and ordinal sums of finite ordinals are equal. For a shorter proof using ax-rep 5237, see nnadjuALT 10181. (Contributed by Paul Chapman, 11-Apr-2009.) (Revised by Mario Carneiro, 6-Feb-2013.) Avoid ax-rep 5237. (Revised by BTernaryTau, 2-Jul-2024.)
⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (card‘(𝐴 ⊔ 𝐵)) = (𝐴 +o 𝐵))
Theorem	nnadjuALT 10181	Shorter proof of nnadju 10180 using ax-rep 5237. (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 10182	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 10183	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 5237. (Revised by BTernaryTau, 3-Jul-2024.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin ∧ (𝐴 ∩ 𝐵) = ∅) → (card‘(𝐴 ∪ 𝐵)) = ((card‘𝐴) +o (card‘𝐵)))
Theorem	ficardun2 10184	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 5237. (Revised by BTernaryTau, 3-Jul-2024.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) → (card‘(𝐴 ∪ 𝐵)) ⊆ ((card‘𝐴) +o (card‘𝐵)))
Theorem	pwsdompw 10185*	Lemma for domtriom 10426. 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))
Theorem	unctb 10186	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) → (𝐴 ∪ 𝐵) ≼ ω)
Theorem	infdjuabs 10187	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 ≼ 𝐴 ∧ 𝐵 ≼ 𝐴) → (𝐴 ⊔ 𝐵) ≈ 𝐴)
Theorem	infunabs 10188	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 ≼ 𝐴 ∧ 𝐵 ≼ 𝐴) → (𝐴 ∪ 𝐵) ≈ 𝐴)
Theorem	infdju 10189	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) → (𝐴 ⊔ 𝐵) ≈ (𝐴 ∪ 𝐵))
Theorem	infdif 10190	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 ≼ 𝐴 ∧ 𝐵 ≺ 𝐴) → (𝐴 ∖ 𝐵) ≈ 𝐴)
Theorem	infdif2 10191	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) → ((𝐴 ∖ 𝐵) ≼ 𝐵 ↔ 𝐴 ≼ 𝐵))
Theorem	infxpdom 10192	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 ≼ 𝐴 ∧ 𝐵 ≼ 𝐴) → (𝐴 × 𝐵) ≼ 𝐴)
Theorem	infxpabs 10193	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 ≠ ∅ ∧ 𝐵 ≼ 𝐴)) → (𝐴 × 𝐵) ≈ 𝐴)
Theorem	infunsdom1 10194	The union of two sets that are strictly dominated by the infinite set 𝑋 is also dominated by 𝑋. This version of infunsdom 10195 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) ∧ (𝐴 ≼ 𝐵 ∧ 𝐵 ≺ 𝑋)) → (𝐴 ∪ 𝐵) ≺ 𝑋)
Theorem	infunsdom 10195	The union of two sets that are strictly dominated by the infinite set 𝑋 is also strictly dominated by 𝑋. (Contributed by Mario Carneiro, 3-May-2015.)
⊢ (((𝑋 ∈ dom card ∧ ω ≼ 𝑋) ∧ (𝐴 ≺ 𝑋 ∧ 𝐵 ≺ 𝑋)) → (𝐴 ∪ 𝐵) ≺ 𝑋)
Theorem	infxp 10196	Absorption law for multiplication with an infinite cardinal. Equivalent to Proposition 10.41 of [TakeutiZaring] p. 95. (Contributed by NM, 28-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
⊢ (((𝐴 ∈ dom card ∧ ω ≼ 𝐴) ∧ (𝐵 ∈ dom card ∧ 𝐵 ≠ ∅)) → (𝐴 × 𝐵) ≈ (𝐴 ∪ 𝐵))
Theorem	pwdjudom 10197	A property of dominance over a powerset, and a main lemma for gchac 10665. Similar to Lemma 2.3 of [KanamoriPincus] p. 420. (Contributed by Mario Carneiro, 15-May-2015.)
⊢ (𝒫 (𝐴 ⊔ 𝐴) ≼ (𝐴 ⊔ 𝐵) → 𝒫 𝐴 ≼ 𝐵)
Theorem	infpss 10198*	Every infinite set has an equinumerous proper subset, proved without AC or Infinity. Exercise 7 of [TakeutiZaring] p. 91. See also infpssALT 10296. (Contributed by NM, 23-Oct-2004.) (Revised by Mario Carneiro, 30-Apr-2015.)
⊢ (ω ≼ 𝐴 → ∃𝑥(𝑥 ⊊ 𝐴 ∧ 𝑥 ≈ 𝐴))
Theorem	infmap2 10199*	An exponentiation law for infinite cardinals. Similar to Lemma 6.2 of [Jech] p. 43. Although this version of infmap 10560 avoids the axiom of choice, it requires the powerset of an infinite set to be well-orderable and so is usually not applicable. (Contributed by NM, 1-Oct-2004.) (Revised by Mario Carneiro, 30-Apr-2015.)
⊢ ((ω ≼ 𝐴 ∧ 𝐵 ≼ 𝐴 ∧ (𝐴 ↑m 𝐵) ∈ dom card) → (𝐴 ↑m 𝐵) ≈ {𝑥 ∣ (𝑥 ⊆ 𝐴 ∧ 𝑥 ≈ 𝐵)})
2.6.15  The Ackermann bijection
Theorem	ackbij2lem1 10200	Lemma for ackbij2 10224. (Contributed by Stefan O'Rear, 18-Nov-2014.)
⊢ (𝐴 ∈ ω → 𝒫 𝐴 ⊆ (𝒫 ω ∩ Fin))