P. 69 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 6801-6900   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	reuen1 6801*	Two ways to express "exactly one". (Contributed by Stefan O'Rear, 28-Oct-2014.)
⊢ (∃!𝑥 ∈ 𝐴 𝜑 ↔ {𝑥 ∈ 𝐴 ∣ 𝜑} ≈ 1o)
Theorem	euen1 6802	Two ways to express "exactly one". (Contributed by Stefan O'Rear, 28-Oct-2014.)
⊢ (∃!𝑥𝜑 ↔ {𝑥 ∣ 𝜑} ≈ 1o)
Theorem	euen1b 6803*	Two ways to express "𝐴 has a unique element". (Contributed by Mario Carneiro, 9-Apr-2015.)
⊢ (𝐴 ≈ 1o ↔ ∃!𝑥 𝑥 ∈ 𝐴)
Theorem	en1uniel 6804	A singleton contains its sole element. (Contributed by Stefan O'Rear, 16-Aug-2015.)
⊢ (𝑆 ≈ 1o → ∪ 𝑆 ∈ 𝑆)
Theorem	2dom 6805*	A set that dominates ordinal 2 has at least 2 different members. (Contributed by NM, 25-Jul-2004.)
⊢ (2o ≼ 𝐴 → ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐴 ¬ 𝑥 = 𝑦)
Theorem	fundmen 6806	A function is equinumerous to its domain. Exercise 4 of [Suppes] p. 98. (Contributed by NM, 28-Jul-2004.) (Revised by Mario Carneiro, 15-Nov-2014.)
⊢ 𝐹 ∈ V    ⇒   ⊢ (Fun 𝐹 → dom 𝐹 ≈ 𝐹)
Theorem	fundmeng 6807	A function is equinumerous to its domain. Exercise 4 of [Suppes] p. 98. (Contributed by NM, 17-Sep-2013.)
⊢ ((𝐹 ∈ 𝑉 ∧ Fun 𝐹) → dom 𝐹 ≈ 𝐹)
Theorem	cnven 6808	A relational set is equinumerous to its converse. (Contributed by Mario Carneiro, 28-Dec-2014.)
⊢ ((Rel 𝐴 ∧ 𝐴 ∈ 𝑉) → 𝐴 ≈ ◡𝐴)
Theorem	cnvct 6809	If a set is dominated by ω, so is its converse. (Contributed by Thierry Arnoux, 29-Dec-2016.)
⊢ (𝐴 ≼ ω → ◡𝐴 ≼ ω)
Theorem	fndmeng 6810	A function is equinumerate to its domain. (Contributed by Paul Chapman, 22-Jun-2011.)
⊢ ((𝐹 Fn 𝐴 ∧ 𝐴 ∈ 𝐶) → 𝐴 ≈ 𝐹)
Theorem	mapsnen 6811	Set exponentiation to a singleton exponent is equinumerous to its base. Exercise 4.43 of [Mendelson] p. 255. (Contributed by NM, 17-Dec-2003.) (Revised by Mario Carneiro, 15-Nov-2014.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ (𝐴 ↑𝑚 {𝐵}) ≈ 𝐴
Theorem	map1 6812	Set exponentiation: ordinal 1 to any set is equinumerous to ordinal 1. Exercise 4.42(b) of [Mendelson] p. 255. (Contributed by NM, 17-Dec-2003.)
⊢ (𝐴 ∈ 𝑉 → (1o ↑𝑚 𝐴) ≈ 1o)
Theorem	en2sn 6813	Two singletons are equinumerous. (Contributed by NM, 9-Nov-2003.)
⊢ ((𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷) → {𝐴} ≈ {𝐵})
Theorem	snfig 6814	A singleton is finite. For the proper class case, see snprc 3658. (Contributed by Jim Kingdon, 13-Apr-2020.)
⊢ (𝐴 ∈ 𝑉 → {𝐴} ∈ Fin)
Theorem	fiprc 6815	The class of finite sets is a proper class. (Contributed by Jeff Hankins, 3-Oct-2008.)
⊢ Fin ∉ V
Theorem	unen 6816	Equinumerosity of union of disjoint sets. Theorem 4 of [Suppes] p. 92. (Contributed by NM, 11-Jun-1998.) (Revised by Mario Carneiro, 26-Apr-2015.)
⊢ (((𝐴 ≈ 𝐵 ∧ 𝐶 ≈ 𝐷) ∧ ((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐷) = ∅)) → (𝐴 ∪ 𝐶) ≈ (𝐵 ∪ 𝐷))
Theorem	enpr2d 6817	A pair with distinct elements is equinumerous to ordinal two. (Contributed by Rohan Ridenour, 3-Aug-2023.)
⊢ (𝜑 → 𝐴 ∈ 𝐶)    &   ⊢ (𝜑 → 𝐵 ∈ 𝐷)    &   ⊢ (𝜑 → ¬ 𝐴 = 𝐵)    ⇒   ⊢ (𝜑 → {𝐴, 𝐵} ≈ 2o)
Theorem	ssct 6818	A subset of a set dominated by ω is dominated by ω. (Contributed by Thierry Arnoux, 31-Jan-2017.)
⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ≼ ω) → 𝐴 ≼ ω)
Theorem	1domsn 6819	A singleton (whether of a set or a proper class) is dominated by one. (Contributed by Jim Kingdon, 1-Mar-2022.)
⊢ {𝐴} ≼ 1o
Theorem	enm 6820*	A set equinumerous to an inhabited set is inhabited. (Contributed by Jim Kingdon, 19-May-2020.)
⊢ ((𝐴 ≈ 𝐵 ∧ ∃𝑥 𝑥 ∈ 𝐴) → ∃𝑦 𝑦 ∈ 𝐵)
Theorem	xpsnen 6821	A set is equinumerous to its Cartesian product with a singleton. Proposition 4.22(c) of [Mendelson] p. 254. (Contributed by NM, 4-Jan-2004.) (Revised by Mario Carneiro, 15-Nov-2014.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ (𝐴 × {𝐵}) ≈ 𝐴
Theorem	xpsneng 6822	A set is equinumerous to its Cartesian product with a singleton. Proposition 4.22(c) of [Mendelson] p. 254. (Contributed by NM, 22-Oct-2004.)
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (𝐴 × {𝐵}) ≈ 𝐴)
Theorem	xp1en 6823	One times a cardinal number. (Contributed by NM, 27-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
⊢ (𝐴 ∈ 𝑉 → (𝐴 × 1o) ≈ 𝐴)
Theorem	endisj 6824*	Any two sets are equinumerous to disjoint sets. Exercise 4.39 of [Mendelson] p. 255. (Contributed by NM, 16-Apr-2004.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ ∃𝑥∃𝑦((𝑥 ≈ 𝐴 ∧ 𝑦 ≈ 𝐵) ∧ (𝑥 ∩ 𝑦) = ∅)
Theorem	xpcomf1o 6825*	The canonical bijection from (𝐴 × 𝐵) to (𝐵 × 𝐴). (Contributed by Mario Carneiro, 23-Apr-2014.)
⊢ 𝐹 = (𝑥 ∈ (𝐴 × 𝐵) ↦ ∪ ◡{𝑥})    ⇒   ⊢ 𝐹:(𝐴 × 𝐵)–1-1-onto→(𝐵 × 𝐴)
Theorem	xpcomco 6826*	Composition with the bijection of xpcomf1o 6825 swaps the arguments to a mapping. (Contributed by Mario Carneiro, 30-May-2015.)
⊢ 𝐹 = (𝑥 ∈ (𝐴 × 𝐵) ↦ ∪ ◡{𝑥})    &   ⊢ 𝐺 = (𝑦 ∈ 𝐵, 𝑧 ∈ 𝐴 ↦ 𝐶)    ⇒   ⊢ (𝐺 ∘ 𝐹) = (𝑧 ∈ 𝐴, 𝑦 ∈ 𝐵 ↦ 𝐶)
Theorem	xpcomen 6827	Commutative law for equinumerosity of Cartesian product. Proposition 4.22(d) of [Mendelson] p. 254. (Contributed by NM, 5-Jan-2004.) (Revised by Mario Carneiro, 15-Nov-2014.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ (𝐴 × 𝐵) ≈ (𝐵 × 𝐴)
Theorem	xpcomeng 6828	Commutative law for equinumerosity of Cartesian product. Proposition 4.22(d) of [Mendelson] p. 254. (Contributed by NM, 27-Mar-2006.)
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (𝐴 × 𝐵) ≈ (𝐵 × 𝐴))
Theorem	xpsnen2g 6829	A set is equinumerous to its Cartesian product with a singleton on the left. (Contributed by Stefan O'Rear, 21-Nov-2014.)
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → ({𝐴} × 𝐵) ≈ 𝐵)
Theorem	xpassen 6830	Associative law for equinumerosity of Cartesian product. Proposition 4.22(e) of [Mendelson] p. 254. (Contributed by NM, 22-Jan-2004.) (Revised by Mario Carneiro, 15-Nov-2014.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    &   ⊢ 𝐶 ∈ V    ⇒   ⊢ ((𝐴 × 𝐵) × 𝐶) ≈ (𝐴 × (𝐵 × 𝐶))
Theorem	xpdom2 6831	Dominance law for Cartesian product. Proposition 10.33(2) of [TakeutiZaring] p. 92. (Contributed by NM, 24-Jul-2004.) (Revised by Mario Carneiro, 15-Nov-2014.)
⊢ 𝐶 ∈ V    ⇒   ⊢ (𝐴 ≼ 𝐵 → (𝐶 × 𝐴) ≼ (𝐶 × 𝐵))
Theorem	xpdom2g 6832	Dominance law for Cartesian product. Theorem 6L(c) of [Enderton] p. 149. (Contributed by Mario Carneiro, 26-Apr-2015.)
⊢ ((𝐶 ∈ 𝑉 ∧ 𝐴 ≼ 𝐵) → (𝐶 × 𝐴) ≼ (𝐶 × 𝐵))
Theorem	xpdom1g 6833	Dominance law for Cartesian product. Theorem 6L(c) of [Enderton] p. 149. (Contributed by NM, 25-Mar-2006.) (Revised by Mario Carneiro, 26-Apr-2015.)
⊢ ((𝐶 ∈ 𝑉 ∧ 𝐴 ≼ 𝐵) → (𝐴 × 𝐶) ≼ (𝐵 × 𝐶))
Theorem	xpdom3m 6834*	A set is dominated by its Cartesian product with an inhabited set. Exercise 6 of [Suppes] p. 98. (Contributed by Jim Kingdon, 15-Apr-2020.)
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ ∃𝑥 𝑥 ∈ 𝐵) → 𝐴 ≼ (𝐴 × 𝐵))
Theorem	xpdom1 6835	Dominance law for Cartesian product. Theorem 6L(c) of [Enderton] p. 149. (Contributed by NM, 28-Sep-2004.) (Revised by NM, 29-Mar-2006.) (Revised by Mario Carneiro, 7-May-2015.)
⊢ 𝐶 ∈ V    ⇒   ⊢ (𝐴 ≼ 𝐵 → (𝐴 × 𝐶) ≼ (𝐵 × 𝐶))
Theorem	fopwdom 6836	Covering implies injection on power sets. (Contributed by Stefan O'Rear, 6-Nov-2014.) (Revised by Mario Carneiro, 24-Jun-2015.)
⊢ ((𝐹 ∈ V ∧ 𝐹:𝐴–onto→𝐵) → 𝒫 𝐵 ≼ 𝒫 𝐴)
Theorem	0domg 6837	Any set dominates the empty set. (Contributed by NM, 26-Oct-2003.) (Revised by Mario Carneiro, 26-Apr-2015.)
⊢ (𝐴 ∈ 𝑉 → ∅ ≼ 𝐴)
Theorem	dom0 6838	A set dominated by the empty set is empty. (Contributed by NM, 22-Nov-2004.)
⊢ (𝐴 ≼ ∅ ↔ 𝐴 = ∅)
Theorem	0dom 6839	Any set dominates the empty set. (Contributed by NM, 26-Oct-2003.) (Revised by Mario Carneiro, 26-Apr-2015.)
⊢ 𝐴 ∈ V    ⇒   ⊢ ∅ ≼ 𝐴
Theorem	enen1 6840	Equality-like theorem for equinumerosity. (Contributed by NM, 18-Dec-2003.)
⊢ (𝐴 ≈ 𝐵 → (𝐴 ≈ 𝐶 ↔ 𝐵 ≈ 𝐶))
Theorem	enen2 6841	Equality-like theorem for equinumerosity. (Contributed by NM, 18-Dec-2003.)
⊢ (𝐴 ≈ 𝐵 → (𝐶 ≈ 𝐴 ↔ 𝐶 ≈ 𝐵))
Theorem	domen1 6842	Equality-like theorem for equinumerosity and dominance. (Contributed by NM, 8-Nov-2003.)
⊢ (𝐴 ≈ 𝐵 → (𝐴 ≼ 𝐶 ↔ 𝐵 ≼ 𝐶))
Theorem	domen2 6843	Equality-like theorem for equinumerosity and dominance. (Contributed by NM, 8-Nov-2003.)
⊢ (𝐴 ≈ 𝐵 → (𝐶 ≼ 𝐴 ↔ 𝐶 ≼ 𝐵))
2.6.29  Equinumerosity (cont.)
Theorem	xpf1o 6844*	Construct a bijection on a Cartesian product given bijections on the factors. (Contributed by Mario Carneiro, 30-May-2015.)
⊢ (𝜑 → (𝑥 ∈ 𝐴 ↦ 𝑋):𝐴–1-1-onto→𝐵)    &   ⊢ (𝜑 → (𝑦 ∈ 𝐶 ↦ 𝑌):𝐶–1-1-onto→𝐷)    ⇒   ⊢ (𝜑 → (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐶 ↦ ⟨𝑋, 𝑌⟩):(𝐴 × 𝐶)–1-1-onto→(𝐵 × 𝐷))
Theorem	xpen 6845	Equinumerosity law for Cartesian product. Proposition 4.22(b) of [Mendelson] p. 254. (Contributed by NM, 24-Jul-2004.)
⊢ ((𝐴 ≈ 𝐵 ∧ 𝐶 ≈ 𝐷) → (𝐴 × 𝐶) ≈ (𝐵 × 𝐷))
Theorem	mapen 6846	Two set exponentiations are equinumerous when their bases and exponents are equinumerous. Theorem 6H(c) of [Enderton] p. 139. (Contributed by NM, 16-Dec-2003.) (Proof shortened by Mario Carneiro, 26-Apr-2015.)
⊢ ((𝐴 ≈ 𝐵 ∧ 𝐶 ≈ 𝐷) → (𝐴 ↑𝑚 𝐶) ≈ (𝐵 ↑𝑚 𝐷))
Theorem	mapdom1g 6847	Order-preserving property of set exponentiation. (Contributed by Jim Kingdon, 15-Jul-2022.)
⊢ ((𝐴 ≼ 𝐵 ∧ 𝐶 ∈ 𝑉) → (𝐴 ↑𝑚 𝐶) ≼ (𝐵 ↑𝑚 𝐶))
Theorem	mapxpen 6848	Equinumerosity law for double set exponentiation. Proposition 10.45 of [TakeutiZaring] p. 96. (Contributed by NM, 21-Feb-2004.) (Revised by Mario Carneiro, 24-Jun-2015.)
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝐶 ∈ 𝑋) → ((𝐴 ↑𝑚 𝐵) ↑𝑚 𝐶) ≈ (𝐴 ↑𝑚 (𝐵 × 𝐶)))
Theorem	xpmapenlem 6849*	Lemma for xpmapen 6850. (Contributed by NM, 1-May-2004.) (Revised by Mario Carneiro, 16-Nov-2014.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    &   ⊢ 𝐶 ∈ V    &   ⊢ 𝐷 = (𝑧 ∈ 𝐶 ↦ (1st ‘(𝑥‘𝑧)))    &   ⊢ 𝑅 = (𝑧 ∈ 𝐶 ↦ (2nd ‘(𝑥‘𝑧)))    &   ⊢ 𝑆 = (𝑧 ∈ 𝐶 ↦ ⟨((1st ‘𝑦)‘𝑧), ((2nd ‘𝑦)‘𝑧)⟩)    ⇒   ⊢ ((𝐴 × 𝐵) ↑𝑚 𝐶) ≈ ((𝐴 ↑𝑚 𝐶) × (𝐵 ↑𝑚 𝐶))
Theorem	xpmapen 6850	Equinumerosity law for set exponentiation of a Cartesian product. Exercise 4.47 of [Mendelson] p. 255. (Contributed by NM, 23-Feb-2004.) (Proof shortened by Mario Carneiro, 16-Nov-2014.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    &   ⊢ 𝐶 ∈ V    ⇒   ⊢ ((𝐴 × 𝐵) ↑𝑚 𝐶) ≈ ((𝐴 ↑𝑚 𝐶) × (𝐵 ↑𝑚 𝐶))
Theorem	ssenen 6851*	Equinumerosity of equinumerous subsets of a set. (Contributed by NM, 30-Sep-2004.) (Revised by Mario Carneiro, 16-Nov-2014.)
⊢ (𝐴 ≈ 𝐵 → {𝑥 ∣ (𝑥 ⊆ 𝐴 ∧ 𝑥 ≈ 𝐶)} ≈ {𝑥 ∣ (𝑥 ⊆ 𝐵 ∧ 𝑥 ≈ 𝐶)})
2.6.30  Pigeonhole Principle
Theorem	phplem1 6852	Lemma for Pigeonhole Principle. If we join a natural number to itself minus an element, we end up with its successor minus the same element. (Contributed by NM, 25-May-1998.)
⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ 𝐴) → ({𝐴} ∪ (𝐴 ∖ {𝐵})) = (suc 𝐴 ∖ {𝐵}))
Theorem	phplem2 6853	Lemma for Pigeonhole Principle. A natural number is equinumerous to its successor minus one of its elements. (Contributed by NM, 11-Jun-1998.) (Revised by Mario Carneiro, 16-Nov-2014.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ 𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵}))
Theorem	phplem3 6854	Lemma for Pigeonhole Principle. A natural number is equinumerous to its successor minus any element of the successor. For a version without the redundant hypotheses, see phplem3g 6856. (Contributed by NM, 26-May-1998.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ suc 𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵}))
Theorem	phplem4 6855	Lemma for Pigeonhole Principle. Equinumerosity of successors implies equinumerosity of the original natural numbers. (Contributed by NM, 28-May-1998.) (Revised by Mario Carneiro, 24-Jun-2015.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (suc 𝐴 ≈ suc 𝐵 → 𝐴 ≈ 𝐵))
Theorem	phplem3g 6856	A natural number is equinumerous to its successor minus any element of the successor. Version of phplem3 6854 with unnecessary hypotheses removed. (Contributed by Jim Kingdon, 1-Sep-2021.)
⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ suc 𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵}))
Theorem	nneneq 6857	Two equinumerous natural numbers are equal. Proposition 10.20 of [TakeutiZaring] p. 90 and its converse. Also compare Corollary 6E of [Enderton] p. 136. (Contributed by NM, 28-May-1998.)
⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (𝐴 ≈ 𝐵 ↔ 𝐴 = 𝐵))
Theorem	php5 6858	A natural number is not equinumerous to its successor. Corollary 10.21(1) of [TakeutiZaring] p. 90. (Contributed by NM, 26-Jul-2004.)
⊢ (𝐴 ∈ ω → ¬ 𝐴 ≈ suc 𝐴)
Theorem	snnen2og 6859	A singleton {𝐴} is never equinumerous with the ordinal number 2. If 𝐴 is a proper class, see snnen2oprc 6860. (Contributed by Jim Kingdon, 1-Sep-2021.)
⊢ (𝐴 ∈ 𝑉 → ¬ {𝐴} ≈ 2o)
Theorem	snnen2oprc 6860	A singleton {𝐴} is never equinumerous with the ordinal number 2. If 𝐴 is a set, see snnen2og 6859. (Contributed by Jim Kingdon, 1-Sep-2021.)
⊢ (¬ 𝐴 ∈ V → ¬ {𝐴} ≈ 2o)
Theorem	1nen2 6861	One and two are not equinumerous. (Contributed by Jim Kingdon, 25-Jan-2022.)
⊢ ¬ 1o ≈ 2o
Theorem	phplem4dom 6862	Dominance of successors implies dominance of the original natural numbers. (Contributed by Jim Kingdon, 1-Sep-2021.)
⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (suc 𝐴 ≼ suc 𝐵 → 𝐴 ≼ 𝐵))
Theorem	php5dom 6863	A natural number does not dominate its successor. (Contributed by Jim Kingdon, 1-Sep-2021.)
⊢ (𝐴 ∈ ω → ¬ suc 𝐴 ≼ 𝐴)
Theorem	nndomo 6864	Cardinal ordering agrees with natural number ordering. Example 3 of [Enderton] p. 146. (Contributed by NM, 17-Jun-1998.)
⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (𝐴 ≼ 𝐵 ↔ 𝐴 ⊆ 𝐵))
Theorem	phpm 6865*	Pigeonhole Principle. A natural number is not equinumerous to a proper subset of itself. By "proper subset" here we mean that there is an element which is in the natural number and not in the subset, or in symbols ∃𝑥𝑥 ∈ (𝐴 ∖ 𝐵) (which is stronger than not being equal in the absence of excluded middle). Theorem (Pigeonhole Principle) of [Enderton] p. 134. The theorem is so-called because you can't put n + 1 pigeons into n holes (if each hole holds only one pigeon). The proof consists of lemmas phplem1 6852 through phplem4 6855, nneneq 6857, and this final piece of the proof. (Contributed by NM, 29-May-1998.)
⊢ ((𝐴 ∈ ω ∧ 𝐵 ⊆ 𝐴 ∧ ∃𝑥 𝑥 ∈ (𝐴 ∖ 𝐵)) → ¬ 𝐴 ≈ 𝐵)
Theorem	phpelm 6866	Pigeonhole Principle. A natural number is not equinumerous to an element of itself. (Contributed by Jim Kingdon, 6-Sep-2021.)
⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ 𝐴) → ¬ 𝐴 ≈ 𝐵)
Theorem	phplem4on 6867	Equinumerosity of successors of an ordinal and a natural number implies equinumerosity of the originals. (Contributed by Jim Kingdon, 5-Sep-2021.)
⊢ ((𝐴 ∈ On ∧ 𝐵 ∈ ω) → (suc 𝐴 ≈ suc 𝐵 → 𝐴 ≈ 𝐵))
2.6.31  Finite sets
Theorem	fict 6868	A finite set is dominated by ω. Also see finct 7115. (Contributed by Thierry Arnoux, 27-Mar-2018.)
⊢ (𝐴 ∈ Fin → 𝐴 ≼ ω)
Theorem	fidceq 6869	Equality of members of a finite set is decidable. This may be counterintuitive: cannot any two sets be elements of a finite set? Well, to show, for example, that {𝐵, 𝐶} is finite would require showing it is equinumerous to 1o or to 2o but to show that you'd need to know 𝐵 = 𝐶 or ¬ 𝐵 = 𝐶, respectively. (Contributed by Jim Kingdon, 5-Sep-2021.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ 𝐴 ∧ 𝐶 ∈ 𝐴) → DECID 𝐵 = 𝐶)
Theorem	fidifsnen 6870	All decrements of a finite set are equinumerous. (Contributed by Jim Kingdon, 9-Sep-2021.)
⊢ ((𝑋 ∈ Fin ∧ 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋) → (𝑋 ∖ {𝐴}) ≈ (𝑋 ∖ {𝐵}))
Theorem	fidifsnid 6871	If we remove a single element from a finite set then put it back in, we end up with the original finite set. This strengthens difsnss 3739 from subset to equality when the set is finite. (Contributed by Jim Kingdon, 9-Sep-2021.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ 𝐴) → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) = 𝐴)
Theorem	nnfi 6872	Natural numbers are finite sets. (Contributed by Stefan O'Rear, 21-Mar-2015.)
⊢ (𝐴 ∈ ω → 𝐴 ∈ Fin)
Theorem	enfi 6873	Equinumerous sets have the same finiteness. (Contributed by NM, 22-Aug-2008.)
⊢ (𝐴 ≈ 𝐵 → (𝐴 ∈ Fin ↔ 𝐵 ∈ Fin))
Theorem	enfii 6874	A set equinumerous to a finite set is finite. (Contributed by Mario Carneiro, 12-Mar-2015.)
⊢ ((𝐵 ∈ Fin ∧ 𝐴 ≈ 𝐵) → 𝐴 ∈ Fin)
Theorem	ssfilem 6875*	Lemma for ssfiexmid 6876. (Contributed by Jim Kingdon, 3-Feb-2022.)
⊢ {𝑧 ∈ {∅} ∣ 𝜑} ∈ Fin    ⇒   ⊢ (𝜑 ∨ ¬ 𝜑)
Theorem	ssfiexmid 6876*	If any subset of a finite set is finite, excluded middle follows. One direction of Theorem 2.1 of [Bauer], p. 485. (Contributed by Jim Kingdon, 19-May-2020.)
⊢ ∀𝑥∀𝑦((𝑥 ∈ Fin ∧ 𝑦 ⊆ 𝑥) → 𝑦 ∈ Fin)    ⇒   ⊢ (𝜑 ∨ ¬ 𝜑)
Theorem	infiexmid 6877*	If the intersection of any finite set and any other set is finite, excluded middle follows. (Contributed by Jim Kingdon, 5-Feb-2022.)
⊢ (𝑥 ∈ Fin → (𝑥 ∩ 𝑦) ∈ Fin)    ⇒   ⊢ (𝜑 ∨ ¬ 𝜑)
Theorem	domfiexmid 6878*	If any set dominated by a finite set is finite, excluded middle follows. (Contributed by Jim Kingdon, 3-Feb-2022.)
⊢ ((𝑥 ∈ Fin ∧ 𝑦 ≼ 𝑥) → 𝑦 ∈ Fin)    ⇒   ⊢ (𝜑 ∨ ¬ 𝜑)
Theorem	dif1en 6879	If a set 𝐴 is equinumerous to the successor of a natural number 𝑀, then 𝐴 with an element removed is equinumerous to 𝑀. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Stefan O'Rear, 16-Aug-2015.)
⊢ ((𝑀 ∈ ω ∧ 𝐴 ≈ suc 𝑀 ∧ 𝑋 ∈ 𝐴) → (𝐴 ∖ {𝑋}) ≈ 𝑀)
Theorem	dif1enen 6880	Subtracting one element from each of two equinumerous finite sets. (Contributed by Jim Kingdon, 5-Jun-2022.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ (𝜑 → 𝐴 ≈ 𝐵)    &   ⊢ (𝜑 → 𝐶 ∈ 𝐴)    &   ⊢ (𝜑 → 𝐷 ∈ 𝐵)    ⇒   ⊢ (𝜑 → (𝐴 ∖ {𝐶}) ≈ (𝐵 ∖ {𝐷}))
Theorem	fiunsnnn 6881	Adding one element to a finite set which is equinumerous to a natural number. (Contributed by Jim Kingdon, 13-Sep-2021.)
⊢ (((𝐴 ∈ Fin ∧ 𝐵 ∈ (V ∖ 𝐴)) ∧ (𝑁 ∈ ω ∧ 𝐴 ≈ 𝑁)) → (𝐴 ∪ {𝐵}) ≈ suc 𝑁)
Theorem	php5fin 6882	A finite set is not equinumerous to a set which adds one element. (Contributed by Jim Kingdon, 13-Sep-2021.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ (V ∖ 𝐴)) → ¬ 𝐴 ≈ (𝐴 ∪ {𝐵}))
Theorem	fisbth 6883	Schroeder-Bernstein Theorem for finite sets. (Contributed by Jim Kingdon, 12-Sep-2021.)
⊢ (((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) ∧ (𝐴 ≼ 𝐵 ∧ 𝐵 ≼ 𝐴)) → 𝐴 ≈ 𝐵)
Theorem	0fin 6884	The empty set is finite. (Contributed by FL, 14-Jul-2008.)
⊢ ∅ ∈ Fin
Theorem	fin0 6885*	A nonempty finite set has at least one element. (Contributed by Jim Kingdon, 10-Sep-2021.)
⊢ (𝐴 ∈ Fin → (𝐴 ≠ ∅ ↔ ∃𝑥 𝑥 ∈ 𝐴))
Theorem	fin0or 6886*	A finite set is either empty or inhabited. (Contributed by Jim Kingdon, 30-Sep-2021.)
⊢ (𝐴 ∈ Fin → (𝐴 = ∅ ∨ ∃𝑥 𝑥 ∈ 𝐴))
Theorem	diffitest 6887*	If subtracting any set from a finite set gives a finite set, any proposition of the form ¬ 𝜑 is decidable. This is not a proof of full excluded middle, but it is close enough to show we won't be able to prove 𝐴 ∈ Fin → (𝐴 ∖ 𝐵) ∈ Fin. (Contributed by Jim Kingdon, 8-Sep-2021.)
⊢ ∀𝑎 ∈ Fin ∀𝑏(𝑎 ∖ 𝑏) ∈ Fin    ⇒   ⊢ (¬ 𝜑 ∨ ¬ ¬ 𝜑)
Theorem	findcard 6888*	Schema for induction on the cardinality of a finite set. The inductive hypothesis is that the result is true on the given set with any one element removed. The result is then proven to be true for all finite sets. (Contributed by Jeff Madsen, 2-Sep-2009.)
⊢ (𝑥 = ∅ → (𝜑 ↔ 𝜓))    &   ⊢ (𝑥 = (𝑦 ∖ {𝑧}) → (𝜑 ↔ 𝜒))    &   ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜃))    &   ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜏))    &   ⊢ 𝜓    &   ⊢ (𝑦 ∈ Fin → (∀𝑧 ∈ 𝑦 𝜒 → 𝜃))    ⇒   ⊢ (𝐴 ∈ Fin → 𝜏)
Theorem	findcard2 6889*	Schema for induction on the cardinality of a finite set. The inductive step shows that the result is true if one more element is added to the set. The result is then proven to be true for all finite sets. (Contributed by Jeff Madsen, 8-Jul-2010.)
⊢ (𝑥 = ∅ → (𝜑 ↔ 𝜓))    &   ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜒))    &   ⊢ (𝑥 = (𝑦 ∪ {𝑧}) → (𝜑 ↔ 𝜃))    &   ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜏))    &   ⊢ 𝜓    &   ⊢ (𝑦 ∈ Fin → (𝜒 → 𝜃))    ⇒   ⊢ (𝐴 ∈ Fin → 𝜏)
Theorem	findcard2s 6890*	Variation of findcard2 6889 requiring that the element added in the induction step not be a member of the original set. (Contributed by Paul Chapman, 30-Nov-2012.)
⊢ (𝑥 = ∅ → (𝜑 ↔ 𝜓))    &   ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜒))    &   ⊢ (𝑥 = (𝑦 ∪ {𝑧}) → (𝜑 ↔ 𝜃))    &   ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜏))    &   ⊢ 𝜓    &   ⊢ ((𝑦 ∈ Fin ∧ ¬ 𝑧 ∈ 𝑦) → (𝜒 → 𝜃))    ⇒   ⊢ (𝐴 ∈ Fin → 𝜏)
Theorem	findcard2d 6891*	Deduction version of findcard2 6889. If you also need 𝑦 ∈ Fin (which doesn't come for free due to ssfiexmid 6876), use findcard2sd 6892 instead. (Contributed by SO, 16-Jul-2018.)
⊢ (𝑥 = ∅ → (𝜓 ↔ 𝜒))    &   ⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜃))    &   ⊢ (𝑥 = (𝑦 ∪ {𝑧}) → (𝜓 ↔ 𝜏))    &   ⊢ (𝑥 = 𝐴 → (𝜓 ↔ 𝜂))    &   ⊢ (𝜑 → 𝜒)    &   ⊢ ((𝜑 ∧ (𝑦 ⊆ 𝐴 ∧ 𝑧 ∈ (𝐴 ∖ 𝑦))) → (𝜃 → 𝜏))    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    ⇒   ⊢ (𝜑 → 𝜂)
Theorem	findcard2sd 6892*	Deduction form of finite set induction . (Contributed by Jim Kingdon, 14-Sep-2021.)
⊢ (𝑥 = ∅ → (𝜓 ↔ 𝜒))    &   ⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜃))    &   ⊢ (𝑥 = (𝑦 ∪ {𝑧}) → (𝜓 ↔ 𝜏))    &   ⊢ (𝑥 = 𝐴 → (𝜓 ↔ 𝜂))    &   ⊢ (𝜑 → 𝜒)    &   ⊢ (((𝜑 ∧ 𝑦 ∈ Fin) ∧ (𝑦 ⊆ 𝐴 ∧ 𝑧 ∈ (𝐴 ∖ 𝑦))) → (𝜃 → 𝜏))    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    ⇒   ⊢ (𝜑 → 𝜂)
Theorem	diffisn 6893	Subtracting a singleton from a finite set produces a finite set. (Contributed by Jim Kingdon, 11-Sep-2021.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ 𝐴) → (𝐴 ∖ {𝐵}) ∈ Fin)
Theorem	diffifi 6894	Subtracting one finite set from another produces a finite set. (Contributed by Jim Kingdon, 8-Sep-2021.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin ∧ 𝐵 ⊆ 𝐴) → (𝐴 ∖ 𝐵) ∈ Fin)
Theorem	infnfi 6895	An infinite set is not finite. (Contributed by Jim Kingdon, 20-Feb-2022.)
⊢ (ω ≼ 𝐴 → ¬ 𝐴 ∈ Fin)
Theorem	ominf 6896	The set of natural numbers is not finite. Although we supply this theorem because we can, the more natural way to express "ω is infinite" is ω ≼ ω which is an instance of domrefg 6767. (Contributed by NM, 2-Jun-1998.)
⊢ ¬ ω ∈ Fin
Theorem	isinfinf 6897*	An infinite set contains subsets of arbitrarily large finite cardinality. (Contributed by Jim Kingdon, 15-Jun-2022.)
⊢ (ω ≼ 𝐴 → ∀𝑛 ∈ ω ∃𝑥(𝑥 ⊆ 𝐴 ∧ 𝑥 ≈ 𝑛))
Theorem	ac6sfi 6898*	Existence of a choice function for finite sets. (Contributed by Jeff Hankins, 26-Jun-2009.) (Proof shortened by Mario Carneiro, 29-Jan-2014.)
⊢ (𝑦 = (𝑓‘𝑥) → (𝜑 ↔ 𝜓))    ⇒   ⊢ ((𝐴 ∈ Fin ∧ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝜑) → ∃𝑓(𝑓:𝐴⟶𝐵 ∧ ∀𝑥 ∈ 𝐴 𝜓))
Theorem	tridc 6899*	A trichotomous order is decidable. (Contributed by Jim Kingdon, 5-Sep-2022.)
⊢ (𝜑 → 𝑅 Po 𝐴)    &   ⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥))    &   ⊢ (𝜑 → 𝐵 ∈ 𝐴)    &   ⊢ (𝜑 → 𝐶 ∈ 𝐴)    ⇒   ⊢ (𝜑 → DECID 𝐵𝑅𝐶)
Theorem	fimax2gtrilemstep 6900*	Lemma for fimax2gtri 6901. The induction step. (Contributed by Jim Kingdon, 5-Sep-2022.)
⊢ (𝜑 → 𝑅 Po 𝐴)    &   ⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥))    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ (𝜑 → 𝐴 ≠ ∅)    &   ⊢ (𝜑 → 𝑈 ∈ Fin)    &   ⊢ (𝜑 → 𝑈 ⊆ 𝐴)    &   ⊢ (𝜑 → 𝑍 ∈ 𝐴)    &   ⊢ (𝜑 → 𝑉 ∈ 𝐴)    &   ⊢ (𝜑 → ¬ 𝑉 ∈ 𝑈)    &   ⊢ (𝜑 → ∀𝑦 ∈ 𝑈 ¬ 𝑍𝑅𝑦)    ⇒   ⊢ (𝜑 → ∃𝑥 ∈ 𝐴 ∀𝑦 ∈ (𝑈 ∪ {𝑉}) ¬ 𝑥𝑅𝑦)