P. 67 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 6601-6700   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	idssen 6601	Equality implies equinumerosity. (Contributed by NM, 30-Apr-1998.) (Revised by Mario Carneiro, 15-Nov-2014.)
⊢ I ⊆ ≈
Theorem	ssdomg 6602	A set dominates its subsets. Theorem 16 of [Suppes] p. 94. (Contributed by NM, 19-Jun-1998.) (Revised by Mario Carneiro, 24-Jun-2015.)
⊢ (𝐵 ∈ 𝑉 → (𝐴 ⊆ 𝐵 → 𝐴 ≼ 𝐵))
Theorem	ener 6603	Equinumerosity is an equivalence relation. (Contributed by NM, 19-Mar-1998.) (Revised by Mario Carneiro, 15-Nov-2014.)
⊢ ≈ Er V
Theorem	ensymb 6604	Symmetry of equinumerosity. Theorem 2 of [Suppes] p. 92. (Contributed by Mario Carneiro, 26-Apr-2015.)
⊢ (𝐴 ≈ 𝐵 ↔ 𝐵 ≈ 𝐴)
Theorem	ensym 6605	Symmetry of equinumerosity. Theorem 2 of [Suppes] p. 92. (Contributed by NM, 26-Oct-2003.) (Revised by Mario Carneiro, 26-Apr-2015.)
⊢ (𝐴 ≈ 𝐵 → 𝐵 ≈ 𝐴)
Theorem	ensymi 6606	Symmetry of equinumerosity. Theorem 2 of [Suppes] p. 92. (Contributed by NM, 25-Sep-2004.)
⊢ 𝐴 ≈ 𝐵    ⇒   ⊢ 𝐵 ≈ 𝐴
Theorem	ensymd 6607	Symmetry of equinumerosity. Deduction form of ensym 6605. (Contributed by David Moews, 1-May-2017.)
⊢ (𝜑 → 𝐴 ≈ 𝐵)    ⇒   ⊢ (𝜑 → 𝐵 ≈ 𝐴)
Theorem	entr 6608	Transitivity of equinumerosity. Theorem 3 of [Suppes] p. 92. (Contributed by NM, 9-Jun-1998.)
⊢ ((𝐴 ≈ 𝐵 ∧ 𝐵 ≈ 𝐶) → 𝐴 ≈ 𝐶)
Theorem	domtr 6609	Transitivity of dominance relation. Theorem 17 of [Suppes] p. 94. (Contributed by NM, 4-Jun-1998.) (Revised by Mario Carneiro, 15-Nov-2014.)
⊢ ((𝐴 ≼ 𝐵 ∧ 𝐵 ≼ 𝐶) → 𝐴 ≼ 𝐶)
Theorem	entri 6610	A chained equinumerosity inference. (Contributed by NM, 25-Sep-2004.)
⊢ 𝐴 ≈ 𝐵    &   ⊢ 𝐵 ≈ 𝐶    ⇒   ⊢ 𝐴 ≈ 𝐶
Theorem	entr2i 6611	A chained equinumerosity inference. (Contributed by NM, 25-Sep-2004.)
⊢ 𝐴 ≈ 𝐵    &   ⊢ 𝐵 ≈ 𝐶    ⇒   ⊢ 𝐶 ≈ 𝐴
Theorem	entr3i 6612	A chained equinumerosity inference. (Contributed by NM, 25-Sep-2004.)
⊢ 𝐴 ≈ 𝐵    &   ⊢ 𝐴 ≈ 𝐶    ⇒   ⊢ 𝐵 ≈ 𝐶
Theorem	entr4i 6613	A chained equinumerosity inference. (Contributed by NM, 25-Sep-2004.)
⊢ 𝐴 ≈ 𝐵    &   ⊢ 𝐶 ≈ 𝐵    ⇒   ⊢ 𝐴 ≈ 𝐶
Theorem	endomtr 6614	Transitivity of equinumerosity and dominance. (Contributed by NM, 7-Jun-1998.)
⊢ ((𝐴 ≈ 𝐵 ∧ 𝐵 ≼ 𝐶) → 𝐴 ≼ 𝐶)
Theorem	domentr 6615	Transitivity of dominance and equinumerosity. (Contributed by NM, 7-Jun-1998.)
⊢ ((𝐴 ≼ 𝐵 ∧ 𝐵 ≈ 𝐶) → 𝐴 ≼ 𝐶)
Theorem	f1imaeng 6616	A one-to-one function's image under a subset of its domain is equinumerous to the subset. (Contributed by Mario Carneiro, 15-May-2015.)
⊢ ((𝐹:𝐴–1-1→𝐵 ∧ 𝐶 ⊆ 𝐴 ∧ 𝐶 ∈ 𝑉) → (𝐹 “ 𝐶) ≈ 𝐶)
Theorem	f1imaen2g 6617	A one-to-one function's image under a subset of its domain is equinumerous to the subset. (This version of f1imaen 6618 does not need ax-setind 4390.) (Contributed by Mario Carneiro, 16-Nov-2014.) (Revised by Mario Carneiro, 25-Jun-2015.)
⊢ (((𝐹:𝐴–1-1→𝐵 ∧ 𝐵 ∈ 𝑉) ∧ (𝐶 ⊆ 𝐴 ∧ 𝐶 ∈ 𝑉)) → (𝐹 “ 𝐶) ≈ 𝐶)
Theorem	f1imaen 6618	A one-to-one function's image under a subset of its domain is equinumerous to the subset. (Contributed by NM, 30-Sep-2004.)
⊢ 𝐶 ∈ V    ⇒   ⊢ ((𝐹:𝐴–1-1→𝐵 ∧ 𝐶 ⊆ 𝐴) → (𝐹 “ 𝐶) ≈ 𝐶)
Theorem	en0 6619	The empty set is equinumerous only to itself. Exercise 1 of [TakeutiZaring] p. 88. (Contributed by NM, 27-May-1998.)
⊢ (𝐴 ≈ ∅ ↔ 𝐴 = ∅)
Theorem	ensn1 6620	A singleton is equinumerous to ordinal one. (Contributed by NM, 4-Nov-2002.)
⊢ 𝐴 ∈ V    ⇒   ⊢ {𝐴} ≈ 1o
Theorem	ensn1g 6621	A singleton is equinumerous to ordinal one. (Contributed by NM, 23-Apr-2004.)
⊢ (𝐴 ∈ 𝑉 → {𝐴} ≈ 1o)
Theorem	enpr1g 6622	{𝐴, 𝐴} has only one element. (Contributed by FL, 15-Feb-2010.)
⊢ (𝐴 ∈ 𝑉 → {𝐴, 𝐴} ≈ 1o)
Theorem	en1 6623*	A set is equinumerous to ordinal one iff it is a singleton. (Contributed by NM, 25-Jul-2004.)
⊢ (𝐴 ≈ 1o ↔ ∃𝑥 𝐴 = {𝑥})
Theorem	en1bg 6624	A set is equinumerous to ordinal one iff it is a singleton. (Contributed by Jim Kingdon, 13-Apr-2020.)
⊢ (𝐴 ∈ 𝑉 → (𝐴 ≈ 1o ↔ 𝐴 = {∪ 𝐴}))
Theorem	reuen1 6625*	Two ways to express "exactly one". (Contributed by Stefan O'Rear, 28-Oct-2014.)
⊢ (∃!𝑥 ∈ 𝐴 𝜑 ↔ {𝑥 ∈ 𝐴 ∣ 𝜑} ≈ 1o)
Theorem	euen1 6626	Two ways to express "exactly one". (Contributed by Stefan O'Rear, 28-Oct-2014.)
⊢ (∃!𝑥𝜑 ↔ {𝑥 ∣ 𝜑} ≈ 1o)
Theorem	euen1b 6627*	Two ways to express "𝐴 has a unique element". (Contributed by Mario Carneiro, 9-Apr-2015.)
⊢ (𝐴 ≈ 1o ↔ ∃!𝑥 𝑥 ∈ 𝐴)
Theorem	en1uniel 6628	A singleton contains its sole element. (Contributed by Stefan O'Rear, 16-Aug-2015.)
⊢ (𝑆 ≈ 1o → ∪ 𝑆 ∈ 𝑆)
Theorem	2dom 6629*	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 6630	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 6631	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 6632	A relational set is equinumerous to its converse. (Contributed by Mario Carneiro, 28-Dec-2014.)
⊢ ((Rel 𝐴 ∧ 𝐴 ∈ 𝑉) → 𝐴 ≈ ◡𝐴)
Theorem	cnvct 6633	If a set is dominated by ω, so is its converse. (Contributed by Thierry Arnoux, 29-Dec-2016.)
⊢ (𝐴 ≼ ω → ◡𝐴 ≼ ω)
Theorem	fndmeng 6634	A function is equinumerate to its domain. (Contributed by Paul Chapman, 22-Jun-2011.)
⊢ ((𝐹 Fn 𝐴 ∧ 𝐴 ∈ 𝐶) → 𝐴 ≈ 𝐹)
Theorem	mapsnen 6635	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 6636	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 6637	Two singletons are equinumerous. (Contributed by NM, 9-Nov-2003.)
⊢ ((𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷) → {𝐴} ≈ {𝐵})
Theorem	snfig 6638	A singleton is finite. For the proper class case, see snprc 3535. (Contributed by Jim Kingdon, 13-Apr-2020.)
⊢ (𝐴 ∈ 𝑉 → {𝐴} ∈ Fin)
Theorem	fiprc 6639	The class of finite sets is a proper class. (Contributed by Jeff Hankins, 3-Oct-2008.)
⊢ Fin ∉ V
Theorem	unen 6640	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	ssct 6641	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 6642	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 6643*	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 6644	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 6645	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 6646	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 6647*	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 6648*	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 6649*	Composition with the bijection of xpcomf1o 6648 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 6650	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 6651	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 6652	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 6653	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 6654	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 6655	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 6656	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 6657*	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 6658	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 6659	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 6660	Any set dominates the empty set. (Contributed by NM, 26-Oct-2003.) (Revised by Mario Carneiro, 26-Apr-2015.)
⊢ (𝐴 ∈ 𝑉 → ∅ ≼ 𝐴)
Theorem	dom0 6661	A set dominated by the empty set is empty. (Contributed by NM, 22-Nov-2004.)
⊢ (𝐴 ≼ ∅ ↔ 𝐴 = ∅)
Theorem	0dom 6662	Any set dominates the empty set. (Contributed by NM, 26-Oct-2003.) (Revised by Mario Carneiro, 26-Apr-2015.)
⊢ 𝐴 ∈ V    ⇒   ⊢ ∅ ≼ 𝐴
Theorem	enen1 6663	Equality-like theorem for equinumerosity. (Contributed by NM, 18-Dec-2003.)
⊢ (𝐴 ≈ 𝐵 → (𝐴 ≈ 𝐶 ↔ 𝐵 ≈ 𝐶))
Theorem	enen2 6664	Equality-like theorem for equinumerosity. (Contributed by NM, 18-Dec-2003.)
⊢ (𝐴 ≈ 𝐵 → (𝐶 ≈ 𝐴 ↔ 𝐶 ≈ 𝐵))
Theorem	domen1 6665	Equality-like theorem for equinumerosity and dominance. (Contributed by NM, 8-Nov-2003.)
⊢ (𝐴 ≈ 𝐵 → (𝐴 ≼ 𝐶 ↔ 𝐵 ≼ 𝐶))
Theorem	domen2 6666	Equality-like theorem for equinumerosity and dominance. (Contributed by NM, 8-Nov-2003.)
⊢ (𝐴 ≈ 𝐵 → (𝐶 ≼ 𝐴 ↔ 𝐶 ≼ 𝐵))
2.6.28  Equinumerosity (cont.)
Theorem	xpf1o 6667*	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 6668	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 6669	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 6670	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 6671	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 6672*	Lemma for xpmapen 6673. (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 6673	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 6674*	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)})
2.6.29  Pigeonhole Principle
Theorem	phplem1 6675	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 6676	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 6677	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 6679. (Contributed by NM, 26-May-1998.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ suc 𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵}))
Theorem	phplem4 6678	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 6679	A natural number is equinumerous to its successor minus any element of the successor. Version of phplem3 6677 with unnecessary hypotheses removed. (Contributed by Jim Kingdon, 1-Sep-2021.)
⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ suc 𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵}))
Theorem	nneneq 6680	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 6681	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 6682	A singleton {𝐴} is never equinumerous with the ordinal number 2. If 𝐴 is a proper class, see snnen2oprc 6683. (Contributed by Jim Kingdon, 1-Sep-2021.)
⊢ (𝐴 ∈ 𝑉 → ¬ {𝐴} ≈ 2o)
Theorem	snnen2oprc 6683	A singleton {𝐴} is never equinumerous with the ordinal number 2. If 𝐴 is a set, see snnen2og 6682. (Contributed by Jim Kingdon, 1-Sep-2021.)
⊢ (¬ 𝐴 ∈ V → ¬ {𝐴} ≈ 2o)
Theorem	1nen2 6684	One and two are not equinumerous. (Contributed by Jim Kingdon, 25-Jan-2022.)
⊢ ¬ 1o ≈ 2o
Theorem	phplem4dom 6685	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 6686	A natural number does not dominate its successor. (Contributed by Jim Kingdon, 1-Sep-2021.)
⊢ (𝐴 ∈ ω → ¬ suc 𝐴 ≼ 𝐴)
Theorem	nndomo 6687	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 6688*	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 6675 through phplem4 6678, nneneq 6680, 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 6689	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 6690	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))
2.6.30  Finite sets
Theorem	fict 6691	A finite set is dominated by ω. Also see finct 6915. (Contributed by Thierry Arnoux, 27-Mar-2018.)
⊢ (𝐴 ∈ Fin → 𝐴 ≼ ω)
Theorem	fidceq 6692	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 6693	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 6694	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 3613 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 6695	Natural numbers are finite sets. (Contributed by Stefan O'Rear, 21-Mar-2015.)
⊢ (𝐴 ∈ ω → 𝐴 ∈ Fin)
Theorem	enfi 6696	Equinumerous sets have the same finiteness. (Contributed by NM, 22-Aug-2008.)
⊢ (𝐴 ≈ 𝐵 → (𝐴 ∈ Fin ↔ 𝐵 ∈ Fin))
Theorem	enfii 6697	A set equinumerous to a finite set is finite. (Contributed by Mario Carneiro, 12-Mar-2015.)
⊢ ((𝐵 ∈ Fin ∧ 𝐴 ≈ 𝐵) → 𝐴 ∈ Fin)
Theorem	ssfilem 6698*	Lemma for ssfiexmid 6699. (Contributed by Jim Kingdon, 3-Feb-2022.)
⊢ {𝑧 ∈ {∅} ∣ 𝜑} ∈ Fin    ⇒   ⊢ (𝜑 ∨ ¬ 𝜑)
Theorem	ssfiexmid 6699*	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 6700*	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)