P. 70 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 6901-7000   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	euen1 6901	Two ways to express "exactly one". (Contributed by Stefan O'Rear, 28-Oct-2014.)
⊢ (∃!𝑥𝜑 ↔ {𝑥 ∣ 𝜑} ≈ 1o)
Theorem	euen1b 6902*	Two ways to express "𝐴 has a unique element". (Contributed by Mario Carneiro, 9-Apr-2015.)
⊢ (𝐴 ≈ 1o ↔ ∃!𝑥 𝑥 ∈ 𝐴)
Theorem	en1uniel 6903	A singleton contains its sole element. (Contributed by Stefan O'Rear, 16-Aug-2015.)
⊢ (𝑆 ≈ 1o → ∪ 𝑆 ∈ 𝑆)
Theorem	2dom 6904*	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 6905	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 6906	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 6907	A relational set is equinumerous to its converse. (Contributed by Mario Carneiro, 28-Dec-2014.)
⊢ ((Rel 𝐴 ∧ 𝐴 ∈ 𝑉) → 𝐴 ≈ ◡𝐴)
Theorem	cnvct 6908	If a set is dominated by ω, so is its converse. (Contributed by Thierry Arnoux, 29-Dec-2016.)
⊢ (𝐴 ≼ ω → ◡𝐴 ≼ ω)
Theorem	fndmeng 6909	A function is equinumerate to its domain. (Contributed by Paul Chapman, 22-Jun-2011.)
⊢ ((𝐹 Fn 𝐴 ∧ 𝐴 ∈ 𝐶) → 𝐴 ≈ 𝐹)
Theorem	mapsnen 6910	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 6911	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 6912	Two singletons are equinumerous. (Contributed by NM, 9-Nov-2003.)
⊢ ((𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷) → {𝐴} ≈ {𝐵})
Theorem	snfig 6913	A singleton is finite. For the proper class case, see snprc 3699. (Contributed by Jim Kingdon, 13-Apr-2020.)
⊢ (𝐴 ∈ 𝑉 → {𝐴} ∈ Fin)
Theorem	fiprc 6914	The class of finite sets is a proper class. (Contributed by Jeff Hankins, 3-Oct-2008.)
⊢ Fin ∉ V
Theorem	unen 6915	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	en2prd 6916	Two proper unordered pairs are equinumerous. (Contributed by BTernaryTau, 23-Dec-2024.)
⊢ (𝜑 → 𝐴 ∈ 𝑉)    &   ⊢ (𝜑 → 𝐵 ∈ 𝑊)    &   ⊢ (𝜑 → 𝐶 ∈ 𝑋)    &   ⊢ (𝜑 → 𝐷 ∈ 𝑌)    &   ⊢ (𝜑 → 𝐴 ≠ 𝐵)    &   ⊢ (𝜑 → 𝐶 ≠ 𝐷)    ⇒   ⊢ (𝜑 → {𝐴, 𝐵} ≈ {𝐶, 𝐷})
Theorem	rex2dom 6917*	A set that has at least 2 different members dominates ordinal 2. (Contributed by BTernaryTau, 30-Dec-2024.)
⊢ ((𝐴 ∈ 𝑉 ∧ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐴 𝑥 ≠ 𝑦) → 2o ≼ 𝐴)
Theorem	enpr2d 6918	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	en2 6919*	A set equinumerous to ordinal 2 is an unordered pair. (Contributed by Mario Carneiro, 5-Jan-2016.)
⊢ (𝐴 ≈ 2o → ∃𝑥∃𝑦 𝐴 = {𝑥, 𝑦})
Theorem	ssct 6920	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 6921	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 6922*	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 6923	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 6924	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 6925	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 6926*	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 6927*	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 6928*	Composition with the bijection of xpcomf1o 6927 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 6929	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 6930	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 6931	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 6932	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 6933	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 6934	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 6935	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 6936*	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 6937	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	pw2f1odclem 6938*	Lemma for pw2f1odc 6939. (Contributed by Mario Carneiro, 6-Oct-2014.)
⊢ (𝜑 → 𝐴 ∈ 𝑉)    &   ⊢ (𝜑 → 𝐵 ∈ 𝑊)    &   ⊢ (𝜑 → 𝐶 ∈ 𝑊)    &   ⊢ (𝜑 → 𝐵 ≠ 𝐶)    &   ⊢ (𝜑 → ∀𝑝 ∈ 𝐴 ∀𝑞 ∈ 𝒫 𝐴DECID 𝑝 ∈ 𝑞)    ⇒   ⊢ (𝜑 → ((𝑆 ∈ 𝒫 𝐴 ∧ 𝐺 = (𝑧 ∈ 𝐴 ↦ if(𝑧 ∈ 𝑆, 𝐶, 𝐵))) ↔ (𝐺 ∈ ({𝐵, 𝐶} ↑𝑚 𝐴) ∧ 𝑆 = (◡𝐺 “ {𝐶}))))
Theorem	pw2f1odc 6939*	The power set of a set is equinumerous to set exponentiation with an unordered pair base of ordinal 2. Generalized from Proposition 10.44 of [TakeutiZaring] p. 96. (Contributed by Mario Carneiro, 6-Oct-2014.)
⊢ (𝜑 → 𝐴 ∈ 𝑉)    &   ⊢ (𝜑 → 𝐵 ∈ 𝑊)    &   ⊢ (𝜑 → 𝐶 ∈ 𝑊)    &   ⊢ (𝜑 → 𝐵 ≠ 𝐶)    &   ⊢ (𝜑 → ∀𝑝 ∈ 𝐴 ∀𝑞 ∈ 𝒫 𝐴DECID 𝑝 ∈ 𝑞)    &   ⊢ 𝐹 = (𝑥 ∈ 𝒫 𝐴 ↦ (𝑧 ∈ 𝐴 ↦ if(𝑧 ∈ 𝑥, 𝐶, 𝐵)))    ⇒   ⊢ (𝜑 → 𝐹:𝒫 𝐴–1-1-onto→({𝐵, 𝐶} ↑𝑚 𝐴))
Theorem	fopwdom 6940	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 6941	Any set dominates the empty set. (Contributed by NM, 26-Oct-2003.) (Revised by Mario Carneiro, 26-Apr-2015.)
⊢ (𝐴 ∈ 𝑉 → ∅ ≼ 𝐴)
Theorem	dom0 6942	A set dominated by the empty set is empty. (Contributed by NM, 22-Nov-2004.)
⊢ (𝐴 ≼ ∅ ↔ 𝐴 = ∅)
Theorem	0dom 6943	Any set dominates the empty set. (Contributed by NM, 26-Oct-2003.) (Revised by Mario Carneiro, 26-Apr-2015.)
⊢ 𝐴 ∈ V    ⇒   ⊢ ∅ ≼ 𝐴
Theorem	enen1 6944	Equality-like theorem for equinumerosity. (Contributed by NM, 18-Dec-2003.)
⊢ (𝐴 ≈ 𝐵 → (𝐴 ≈ 𝐶 ↔ 𝐵 ≈ 𝐶))
Theorem	enen2 6945	Equality-like theorem for equinumerosity. (Contributed by NM, 18-Dec-2003.)
⊢ (𝐴 ≈ 𝐵 → (𝐶 ≈ 𝐴 ↔ 𝐶 ≈ 𝐵))
Theorem	domen1 6946	Equality-like theorem for equinumerosity and dominance. (Contributed by NM, 8-Nov-2003.)
⊢ (𝐴 ≈ 𝐵 → (𝐴 ≼ 𝐶 ↔ 𝐵 ≼ 𝐶))
Theorem	domen2 6947	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 6948*	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 6949	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 6950	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 6951	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 6952	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 6953*	Lemma for xpmapen 6954. (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 6954	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 6955*	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 6956	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 6957	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 6958	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 6960. (Contributed by NM, 26-May-1998.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ suc 𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵}))
Theorem	phplem4 6959	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 6960	A natural number is equinumerous to its successor minus any element of the successor. Version of phplem3 6958 with unnecessary hypotheses removed. (Contributed by Jim Kingdon, 1-Sep-2021.)
⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ suc 𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵}))
Theorem	nneneq 6961	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 6962	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 6963	A singleton {𝐴} is never equinumerous with the ordinal number 2. If 𝐴 is a proper class, see snnen2oprc 6964. (Contributed by Jim Kingdon, 1-Sep-2021.)
⊢ (𝐴 ∈ 𝑉 → ¬ {𝐴} ≈ 2o)
Theorem	snnen2oprc 6964	A singleton {𝐴} is never equinumerous with the ordinal number 2. If 𝐴 is a set, see snnen2og 6963. (Contributed by Jim Kingdon, 1-Sep-2021.)
⊢ (¬ 𝐴 ∈ V → ¬ {𝐴} ≈ 2o)
Theorem	1nen2 6965	One and two are not equinumerous. (Contributed by Jim Kingdon, 25-Jan-2022.)
⊢ ¬ 1o ≈ 2o
Theorem	phplem4dom 6966	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 6967	A natural number does not dominate its successor. (Contributed by Jim Kingdon, 1-Sep-2021.)
⊢ (𝐴 ∈ ω → ¬ suc 𝐴 ≼ 𝐴)
Theorem	nndomo 6968	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 6969*	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 6956 through phplem4 6959, nneneq 6961, 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 6970	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 6971	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 6972	A finite set is dominated by ω. Also see finct 7225. (Contributed by Thierry Arnoux, 27-Mar-2018.)
⊢ (𝐴 ∈ Fin → 𝐴 ≼ ω)
Theorem	fidceq 6973	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 6974	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 6975	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 3781 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 6976	Natural numbers are finite sets. (Contributed by Stefan O'Rear, 21-Mar-2015.)
⊢ (𝐴 ∈ ω → 𝐴 ∈ Fin)
Theorem	enfi 6977	Equinumerous sets have the same finiteness. (Contributed by NM, 22-Aug-2008.)
⊢ (𝐴 ≈ 𝐵 → (𝐴 ∈ Fin ↔ 𝐵 ∈ Fin))
Theorem	enfii 6978	A set equinumerous to a finite set is finite. (Contributed by Mario Carneiro, 12-Mar-2015.)
⊢ ((𝐵 ∈ Fin ∧ 𝐴 ≈ 𝐵) → 𝐴 ∈ Fin)
Theorem	ssfilem 6979*	Lemma for ssfiexmid 6980. (Contributed by Jim Kingdon, 3-Feb-2022.)
⊢ {𝑧 ∈ {∅} ∣ 𝜑} ∈ Fin    ⇒   ⊢ (𝜑 ∨ ¬ 𝜑)
Theorem	ssfiexmid 6980*	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 6981*	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 6982*	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 6983	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 6984	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 6985	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 6986	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 6987	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 6988	The empty set is finite. (Contributed by FL, 14-Jul-2008.)
⊢ ∅ ∈ Fin
Theorem	fin0 6989*	A nonempty finite set has at least one element. (Contributed by Jim Kingdon, 10-Sep-2021.)
⊢ (𝐴 ∈ Fin → (𝐴 ≠ ∅ ↔ ∃𝑥 𝑥 ∈ 𝐴))
Theorem	fin0or 6990*	A finite set is either empty or inhabited. (Contributed by Jim Kingdon, 30-Sep-2021.)
⊢ (𝐴 ∈ Fin → (𝐴 = ∅ ∨ ∃𝑥 𝑥 ∈ 𝐴))
Theorem	diffitest 6991*	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 6992*	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 6993*	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 6994*	Variation of findcard2 6993 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 6995*	Deduction version of findcard2 6993. If you also need 𝑦 ∈ Fin (which doesn't come for free due to ssfiexmid 6980), use findcard2sd 6996 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 6996*	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 6997	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 6998	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 6999	An infinite set is not finite. (Contributed by Jim Kingdon, 20-Feb-2022.)
⊢ (ω ≼ 𝐴 → ¬ 𝐴 ∈ Fin)
Theorem	ominf 7000	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 6865. (Contributed by NM, 2-Jun-1998.)
⊢ ¬ ω ∈ Fin