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Theorem List for Intuitionistic Logic Explorer - 6801-6900   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremdom3d 6801* A mapping (first hypothesis) that is one-to-one (second hypothesis) implies its domain is dominated by its codomain. (Contributed by Mario Carneiro, 20-May-2013.)
(𝜑 → (𝑥𝐴𝐶𝐵))    &   (𝜑 → ((𝑥𝐴𝑦𝐴) → (𝐶 = 𝐷𝑥 = 𝑦)))    &   (𝜑𝐴𝑉)    &   (𝜑𝐵𝑊)       (𝜑𝐴𝐵)
 
Theoremdom2 6802* A mapping (first hypothesis) that is one-to-one (second hypothesis) implies its domain is dominated by its codomain. 𝐶 and 𝐷 can be read 𝐶(𝑥) and 𝐷(𝑦), as can be inferred from their distinct variable conditions. (Contributed by NM, 26-Oct-2003.)
(𝑥𝐴𝐶𝐵)    &   ((𝑥𝐴𝑦𝐴) → (𝐶 = 𝐷𝑥 = 𝑦))       (𝐵𝑉𝐴𝐵)
 
Theoremdom3 6803* A mapping (first hypothesis) that is one-to-one (second hypothesis) implies its domain is dominated by its codomain. 𝐶 and 𝐷 can be read 𝐶(𝑥) and 𝐷(𝑦), as can be inferred from their distinct variable conditions. (Contributed by Mario Carneiro, 20-May-2013.)
(𝑥𝐴𝐶𝐵)    &   ((𝑥𝐴𝑦𝐴) → (𝐶 = 𝐷𝑥 = 𝑦))       ((𝐴𝑉𝐵𝑊) → 𝐴𝐵)
 
Theoremidssen 6804 Equality implies equinumerosity. (Contributed by NM, 30-Apr-1998.) (Revised by Mario Carneiro, 15-Nov-2014.)
I ⊆ ≈
 
Theoremssdomg 6805 A set dominates its subsets. Theorem 16 of [Suppes] p. 94. (Contributed by NM, 19-Jun-1998.) (Revised by Mario Carneiro, 24-Jun-2015.)
(𝐵𝑉 → (𝐴𝐵𝐴𝐵))
 
Theoremener 6806 Equinumerosity is an equivalence relation. (Contributed by NM, 19-Mar-1998.) (Revised by Mario Carneiro, 15-Nov-2014.)
≈ Er V
 
Theoremensymb 6807 Symmetry of equinumerosity. Theorem 2 of [Suppes] p. 92. (Contributed by Mario Carneiro, 26-Apr-2015.)
(𝐴𝐵𝐵𝐴)
 
Theoremensym 6808 Symmetry of equinumerosity. Theorem 2 of [Suppes] p. 92. (Contributed by NM, 26-Oct-2003.) (Revised by Mario Carneiro, 26-Apr-2015.)
(𝐴𝐵𝐵𝐴)
 
Theoremensymi 6809 Symmetry of equinumerosity. Theorem 2 of [Suppes] p. 92. (Contributed by NM, 25-Sep-2004.)
𝐴𝐵       𝐵𝐴
 
Theoremensymd 6810 Symmetry of equinumerosity. Deduction form of ensym 6808. (Contributed by David Moews, 1-May-2017.)
(𝜑𝐴𝐵)       (𝜑𝐵𝐴)
 
Theorementr 6811 Transitivity of equinumerosity. Theorem 3 of [Suppes] p. 92. (Contributed by NM, 9-Jun-1998.)
((𝐴𝐵𝐵𝐶) → 𝐴𝐶)
 
Theoremdomtr 6812 Transitivity of dominance relation. Theorem 17 of [Suppes] p. 94. (Contributed by NM, 4-Jun-1998.) (Revised by Mario Carneiro, 15-Nov-2014.)
((𝐴𝐵𝐵𝐶) → 𝐴𝐶)
 
Theorementri 6813 A chained equinumerosity inference. (Contributed by NM, 25-Sep-2004.)
𝐴𝐵    &   𝐵𝐶       𝐴𝐶
 
Theorementr2i 6814 A chained equinumerosity inference. (Contributed by NM, 25-Sep-2004.)
𝐴𝐵    &   𝐵𝐶       𝐶𝐴
 
Theorementr3i 6815 A chained equinumerosity inference. (Contributed by NM, 25-Sep-2004.)
𝐴𝐵    &   𝐴𝐶       𝐵𝐶
 
Theorementr4i 6816 A chained equinumerosity inference. (Contributed by NM, 25-Sep-2004.)
𝐴𝐵    &   𝐶𝐵       𝐴𝐶
 
Theoremendomtr 6817 Transitivity of equinumerosity and dominance. (Contributed by NM, 7-Jun-1998.)
((𝐴𝐵𝐵𝐶) → 𝐴𝐶)
 
Theoremdomentr 6818 Transitivity of dominance and equinumerosity. (Contributed by NM, 7-Jun-1998.)
((𝐴𝐵𝐵𝐶) → 𝐴𝐶)
 
Theoremf1imaeng 6819 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:𝐴1-1𝐵𝐶𝐴𝐶𝑉) → (𝐹𝐶) ≈ 𝐶)
 
Theoremf1imaen2g 6820 A one-to-one function's image under a subset of its domain is equinumerous to the subset. (This version of f1imaen 6821 does not need ax-setind 4554.) (Contributed by Mario Carneiro, 16-Nov-2014.) (Revised by Mario Carneiro, 25-Jun-2015.)
(((𝐹:𝐴1-1𝐵𝐵𝑉) ∧ (𝐶𝐴𝐶𝑉)) → (𝐹𝐶) ≈ 𝐶)
 
Theoremf1imaen 6821 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:𝐴1-1𝐵𝐶𝐴) → (𝐹𝐶) ≈ 𝐶)
 
Theoremen0 6822 The empty set is equinumerous only to itself. Exercise 1 of [TakeutiZaring] p. 88. (Contributed by NM, 27-May-1998.)
(𝐴 ≈ ∅ ↔ 𝐴 = ∅)
 
Theoremensn1 6823 A singleton is equinumerous to ordinal one. (Contributed by NM, 4-Nov-2002.)
𝐴 ∈ V       {𝐴} ≈ 1o
 
Theoremensn1g 6824 A singleton is equinumerous to ordinal one. (Contributed by NM, 23-Apr-2004.)
(𝐴𝑉 → {𝐴} ≈ 1o)
 
Theoremenpr1g 6825 {𝐴, 𝐴} has only one element. (Contributed by FL, 15-Feb-2010.)
(𝐴𝑉 → {𝐴, 𝐴} ≈ 1o)
 
Theoremen1 6826* A set is equinumerous to ordinal one iff it is a singleton. (Contributed by NM, 25-Jul-2004.)
(𝐴 ≈ 1o ↔ ∃𝑥 𝐴 = {𝑥})
 
Theoremen1bg 6827 A set is equinumerous to ordinal one iff it is a singleton. (Contributed by Jim Kingdon, 13-Apr-2020.)
(𝐴𝑉 → (𝐴 ≈ 1o𝐴 = { 𝐴}))
 
Theoremreuen1 6828* Two ways to express "exactly one". (Contributed by Stefan O'Rear, 28-Oct-2014.)
(∃!𝑥𝐴 𝜑 ↔ {𝑥𝐴𝜑} ≈ 1o)
 
Theoremeuen1 6829 Two ways to express "exactly one". (Contributed by Stefan O'Rear, 28-Oct-2014.)
(∃!𝑥𝜑 ↔ {𝑥𝜑} ≈ 1o)
 
Theoremeuen1b 6830* Two ways to express "𝐴 has a unique element". (Contributed by Mario Carneiro, 9-Apr-2015.)
(𝐴 ≈ 1o ↔ ∃!𝑥 𝑥𝐴)
 
Theoremen1uniel 6831 A singleton contains its sole element. (Contributed by Stefan O'Rear, 16-Aug-2015.)
(𝑆 ≈ 1o 𝑆𝑆)
 
Theorem2dom 6832* A set that dominates ordinal 2 has at least 2 different members. (Contributed by NM, 25-Jul-2004.)
(2o𝐴 → ∃𝑥𝐴𝑦𝐴 ¬ 𝑥 = 𝑦)
 
Theoremfundmen 6833 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 𝐹𝐹)
 
Theoremfundmeng 6834 A function is equinumerous to its domain. Exercise 4 of [Suppes] p. 98. (Contributed by NM, 17-Sep-2013.)
((𝐹𝑉 ∧ Fun 𝐹) → dom 𝐹𝐹)
 
Theoremcnven 6835 A relational set is equinumerous to its converse. (Contributed by Mario Carneiro, 28-Dec-2014.)
((Rel 𝐴𝐴𝑉) → 𝐴𝐴)
 
Theoremcnvct 6836 If a set is dominated by ω, so is its converse. (Contributed by Thierry Arnoux, 29-Dec-2016.)
(𝐴 ≼ ω → 𝐴 ≼ ω)
 
Theoremfndmeng 6837 A function is equinumerate to its domain. (Contributed by Paul Chapman, 22-Jun-2011.)
((𝐹 Fn 𝐴𝐴𝐶) → 𝐴𝐹)
 
Theoremmapsnen 6838 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       (𝐴𝑚 {𝐵}) ≈ 𝐴
 
Theoremmap1 6839 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.)
(𝐴𝑉 → (1o𝑚 𝐴) ≈ 1o)
 
Theoremen2sn 6840 Two singletons are equinumerous. (Contributed by NM, 9-Nov-2003.)
((𝐴𝐶𝐵𝐷) → {𝐴} ≈ {𝐵})
 
Theoremsnfig 6841 A singleton is finite. For the proper class case, see snprc 3672. (Contributed by Jim Kingdon, 13-Apr-2020.)
(𝐴𝑉 → {𝐴} ∈ Fin)
 
Theoremfiprc 6842 The class of finite sets is a proper class. (Contributed by Jeff Hankins, 3-Oct-2008.)
Fin ∉ V
 
Theoremunen 6843 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.)
(((𝐴𝐵𝐶𝐷) ∧ ((𝐴𝐶) = ∅ ∧ (𝐵𝐷) = ∅)) → (𝐴𝐶) ≈ (𝐵𝐷))
 
Theoremenpr2d 6844 A pair with distinct elements is equinumerous to ordinal two. (Contributed by Rohan Ridenour, 3-Aug-2023.)
(𝜑𝐴𝐶)    &   (𝜑𝐵𝐷)    &   (𝜑 → ¬ 𝐴 = 𝐵)       (𝜑 → {𝐴, 𝐵} ≈ 2o)
 
Theoremssct 6845 A subset of a set dominated by ω is dominated by ω. (Contributed by Thierry Arnoux, 31-Jan-2017.)
((𝐴𝐵𝐵 ≼ ω) → 𝐴 ≼ ω)
 
Theorem1domsn 6846 A singleton (whether of a set or a proper class) is dominated by one. (Contributed by Jim Kingdon, 1-Mar-2022.)
{𝐴} ≼ 1o
 
Theoremenm 6847* A set equinumerous to an inhabited set is inhabited. (Contributed by Jim Kingdon, 19-May-2020.)
((𝐴𝐵 ∧ ∃𝑥 𝑥𝐴) → ∃𝑦 𝑦𝐵)
 
Theoremxpsnen 6848 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
 
Theoremxpsneng 6849 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.)
((𝐴𝑉𝐵𝑊) → (𝐴 × {𝐵}) ≈ 𝐴)
 
Theoremxp1en 6850 One times a cardinal number. (Contributed by NM, 27-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
(𝐴𝑉 → (𝐴 × 1o) ≈ 𝐴)
 
Theoremendisj 6851* 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       𝑥𝑦((𝑥𝐴𝑦𝐵) ∧ (𝑥𝑦) = ∅)
 
Theoremxpcomf1o 6852* 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)
 
Theoremxpcomco 6853* Composition with the bijection of xpcomf1o 6852 swaps the arguments to a mapping. (Contributed by Mario Carneiro, 30-May-2015.)
𝐹 = (𝑥 ∈ (𝐴 × 𝐵) ↦ {𝑥})    &   𝐺 = (𝑦𝐵, 𝑧𝐴𝐶)       (𝐺𝐹) = (𝑧𝐴, 𝑦𝐵𝐶)
 
Theoremxpcomen 6854 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)
 
Theoremxpcomeng 6855 Commutative law for equinumerosity of Cartesian product. Proposition 4.22(d) of [Mendelson] p. 254. (Contributed by NM, 27-Mar-2006.)
((𝐴𝑉𝐵𝑊) → (𝐴 × 𝐵) ≈ (𝐵 × 𝐴))
 
Theoremxpsnen2g 6856 A set is equinumerous to its Cartesian product with a singleton on the left. (Contributed by Stefan O'Rear, 21-Nov-2014.)
((𝐴𝑉𝐵𝑊) → ({𝐴} × 𝐵) ≈ 𝐵)
 
Theoremxpassen 6857 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))
 
Theoremxpdom2 6858 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       (𝐴𝐵 → (𝐶 × 𝐴) ≼ (𝐶 × 𝐵))
 
Theoremxpdom2g 6859 Dominance law for Cartesian product. Theorem 6L(c) of [Enderton] p. 149. (Contributed by Mario Carneiro, 26-Apr-2015.)
((𝐶𝑉𝐴𝐵) → (𝐶 × 𝐴) ≼ (𝐶 × 𝐵))
 
Theoremxpdom1g 6860 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.)
((𝐶𝑉𝐴𝐵) → (𝐴 × 𝐶) ≼ (𝐵 × 𝐶))
 
Theoremxpdom3m 6861* 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.)
((𝐴𝑉𝐵𝑊 ∧ ∃𝑥 𝑥𝐵) → 𝐴 ≼ (𝐴 × 𝐵))
 
Theoremxpdom1 6862 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       (𝐴𝐵 → (𝐴 × 𝐶) ≼ (𝐵 × 𝐶))
 
Theorempw2f1odclem 6863* Lemma for pw2f1odc 6864. (Contributed by Mario Carneiro, 6-Oct-2014.)
(𝜑𝐴𝑉)    &   (𝜑𝐵𝑊)    &   (𝜑𝐶𝑊)    &   (𝜑𝐵𝐶)    &   (𝜑 → ∀𝑝𝐴𝑞 ∈ 𝒫 𝐴DECID 𝑝𝑞)       (𝜑 → ((𝑆 ∈ 𝒫 𝐴𝐺 = (𝑧𝐴 ↦ if(𝑧𝑆, 𝐶, 𝐵))) ↔ (𝐺 ∈ ({𝐵, 𝐶} ↑𝑚 𝐴) ∧ 𝑆 = (𝐺 “ {𝐶}))))
 
Theorempw2f1odc 6864* 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.)
(𝜑𝐴𝑉)    &   (𝜑𝐵𝑊)    &   (𝜑𝐶𝑊)    &   (𝜑𝐵𝐶)    &   (𝜑 → ∀𝑝𝐴𝑞 ∈ 𝒫 𝐴DECID 𝑝𝑞)    &   𝐹 = (𝑥 ∈ 𝒫 𝐴 ↦ (𝑧𝐴 ↦ if(𝑧𝑥, 𝐶, 𝐵)))       (𝜑𝐹:𝒫 𝐴1-1-onto→({𝐵, 𝐶} ↑𝑚 𝐴))
 
Theoremfopwdom 6865 Covering implies injection on power sets. (Contributed by Stefan O'Rear, 6-Nov-2014.) (Revised by Mario Carneiro, 24-Jun-2015.)
((𝐹 ∈ V ∧ 𝐹:𝐴onto𝐵) → 𝒫 𝐵 ≼ 𝒫 𝐴)
 
Theorem0domg 6866 Any set dominates the empty set. (Contributed by NM, 26-Oct-2003.) (Revised by Mario Carneiro, 26-Apr-2015.)
(𝐴𝑉 → ∅ ≼ 𝐴)
 
Theoremdom0 6867 A set dominated by the empty set is empty. (Contributed by NM, 22-Nov-2004.)
(𝐴 ≼ ∅ ↔ 𝐴 = ∅)
 
Theorem0dom 6868 Any set dominates the empty set. (Contributed by NM, 26-Oct-2003.) (Revised by Mario Carneiro, 26-Apr-2015.)
𝐴 ∈ V       ∅ ≼ 𝐴
 
Theoremenen1 6869 Equality-like theorem for equinumerosity. (Contributed by NM, 18-Dec-2003.)
(𝐴𝐵 → (𝐴𝐶𝐵𝐶))
 
Theoremenen2 6870 Equality-like theorem for equinumerosity. (Contributed by NM, 18-Dec-2003.)
(𝐴𝐵 → (𝐶𝐴𝐶𝐵))
 
Theoremdomen1 6871 Equality-like theorem for equinumerosity and dominance. (Contributed by NM, 8-Nov-2003.)
(𝐴𝐵 → (𝐴𝐶𝐵𝐶))
 
Theoremdomen2 6872 Equality-like theorem for equinumerosity and dominance. (Contributed by NM, 8-Nov-2003.)
(𝐴𝐵 → (𝐶𝐴𝐶𝐵))
 
2.6.29  Equinumerosity (cont.)
 
Theoremxpf1o 6873* Construct a bijection on a Cartesian product given bijections on the factors. (Contributed by Mario Carneiro, 30-May-2015.)
(𝜑 → (𝑥𝐴𝑋):𝐴1-1-onto𝐵)    &   (𝜑 → (𝑦𝐶𝑌):𝐶1-1-onto𝐷)       (𝜑 → (𝑥𝐴, 𝑦𝐶 ↦ ⟨𝑋, 𝑌⟩):(𝐴 × 𝐶)–1-1-onto→(𝐵 × 𝐷))
 
Theoremxpen 6874 Equinumerosity law for Cartesian product. Proposition 4.22(b) of [Mendelson] p. 254. (Contributed by NM, 24-Jul-2004.)
((𝐴𝐵𝐶𝐷) → (𝐴 × 𝐶) ≈ (𝐵 × 𝐷))
 
Theoremmapen 6875 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.)
((𝐴𝐵𝐶𝐷) → (𝐴𝑚 𝐶) ≈ (𝐵𝑚 𝐷))
 
Theoremmapdom1g 6876 Order-preserving property of set exponentiation. (Contributed by Jim Kingdon, 15-Jul-2022.)
((𝐴𝐵𝐶𝑉) → (𝐴𝑚 𝐶) ≼ (𝐵𝑚 𝐶))
 
Theoremmapxpen 6877 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.)
((𝐴𝑉𝐵𝑊𝐶𝑋) → ((𝐴𝑚 𝐵) ↑𝑚 𝐶) ≈ (𝐴𝑚 (𝐵 × 𝐶)))
 
Theoremxpmapenlem 6878* Lemma for xpmapen 6879. (Contributed by NM, 1-May-2004.) (Revised by Mario Carneiro, 16-Nov-2014.)
𝐴 ∈ V    &   𝐵 ∈ V    &   𝐶 ∈ V    &   𝐷 = (𝑧𝐶 ↦ (1st ‘(𝑥𝑧)))    &   𝑅 = (𝑧𝐶 ↦ (2nd ‘(𝑥𝑧)))    &   𝑆 = (𝑧𝐶 ↦ ⟨((1st𝑦)‘𝑧), ((2nd𝑦)‘𝑧)⟩)       ((𝐴 × 𝐵) ↑𝑚 𝐶) ≈ ((𝐴𝑚 𝐶) × (𝐵𝑚 𝐶))
 
Theoremxpmapen 6879 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) ≈ ((𝐴𝑚 𝐶) × (𝐵𝑚 𝐶))
 
Theoremssenen 6880* Equinumerosity of equinumerous subsets of a set. (Contributed by NM, 30-Sep-2004.) (Revised by Mario Carneiro, 16-Nov-2014.)
(𝐴𝐵 → {𝑥 ∣ (𝑥𝐴𝑥𝐶)} ≈ {𝑥 ∣ (𝑥𝐵𝑥𝐶)})
 
2.6.30  Pigeonhole Principle
 
Theoremphplem1 6881 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 ∧ 𝐵𝐴) → ({𝐴} ∪ (𝐴 ∖ {𝐵})) = (suc 𝐴 ∖ {𝐵}))
 
Theoremphplem2 6882 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 ∧ 𝐵𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵}))
 
Theoremphplem3 6883 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 6885. (Contributed by NM, 26-May-1998.)
𝐴 ∈ V    &   𝐵 ∈ V       ((𝐴 ∈ ω ∧ 𝐵 ∈ suc 𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵}))
 
Theoremphplem4 6884 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 𝐵𝐴𝐵))
 
Theoremphplem3g 6885 A natural number is equinumerous to its successor minus any element of the successor. Version of phplem3 6883 with unnecessary hypotheses removed. (Contributed by Jim Kingdon, 1-Sep-2021.)
((𝐴 ∈ ω ∧ 𝐵 ∈ suc 𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵}))
 
Theoremnneneq 6886 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) → (𝐴𝐵𝐴 = 𝐵))
 
Theoremphp5 6887 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)
 
Theoremsnnen2og 6888 A singleton {𝐴} is never equinumerous with the ordinal number 2. If 𝐴 is a proper class, see snnen2oprc 6889. (Contributed by Jim Kingdon, 1-Sep-2021.)
(𝐴𝑉 → ¬ {𝐴} ≈ 2o)
 
Theoremsnnen2oprc 6889 A singleton {𝐴} is never equinumerous with the ordinal number 2. If 𝐴 is a set, see snnen2og 6888. (Contributed by Jim Kingdon, 1-Sep-2021.)
𝐴 ∈ V → ¬ {𝐴} ≈ 2o)
 
Theorem1nen2 6890 One and two are not equinumerous. (Contributed by Jim Kingdon, 25-Jan-2022.)
¬ 1o ≈ 2o
 
Theoremphplem4dom 6891 Dominance of successors implies dominance of the original natural numbers. (Contributed by Jim Kingdon, 1-Sep-2021.)
((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (suc 𝐴 ≼ suc 𝐵𝐴𝐵))
 
Theoremphp5dom 6892 A natural number does not dominate its successor. (Contributed by Jim Kingdon, 1-Sep-2021.)
(𝐴 ∈ ω → ¬ suc 𝐴𝐴)
 
Theoremnndomo 6893 Cardinal ordering agrees with natural number ordering. Example 3 of [Enderton] p. 146. (Contributed by NM, 17-Jun-1998.)
((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (𝐴𝐵𝐴𝐵))
 
Theoremphpm 6894* 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 ∈ (𝐴𝐵) (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 6881 through phplem4 6884, nneneq 6886, and this final piece of the proof. (Contributed by NM, 29-May-1998.)
((𝐴 ∈ ω ∧ 𝐵𝐴 ∧ ∃𝑥 𝑥 ∈ (𝐴𝐵)) → ¬ 𝐴𝐵)
 
Theoremphpelm 6895 Pigeonhole Principle. A natural number is not equinumerous to an element of itself. (Contributed by Jim Kingdon, 6-Sep-2021.)
((𝐴 ∈ ω ∧ 𝐵𝐴) → ¬ 𝐴𝐵)
 
Theoremphplem4on 6896 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 𝐵𝐴𝐵))
 
2.6.31  Finite sets
 
Theoremfict 6897 A finite set is dominated by ω. Also see finct 7146. (Contributed by Thierry Arnoux, 27-Mar-2018.)
(𝐴 ∈ Fin → 𝐴 ≼ ω)
 
Theoremfidceq 6898 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 ∧ 𝐵𝐴𝐶𝐴) → DECID 𝐵 = 𝐶)
 
Theoremfidifsnen 6899 All decrements of a finite set are equinumerous. (Contributed by Jim Kingdon, 9-Sep-2021.)
((𝑋 ∈ Fin ∧ 𝐴𝑋𝐵𝑋) → (𝑋 ∖ {𝐴}) ≈ (𝑋 ∖ {𝐵}))
 
Theoremfidifsnid 6900 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 3753 from subset to equality when the set is finite. (Contributed by Jim Kingdon, 9-Sep-2021.)
((𝐴 ∈ Fin ∧ 𝐵𝐴) → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) = 𝐴)
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