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Theorem List for Intuitionistic Logic Explorer - 6801-6900   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremfundmeng 6801 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 6802 A relational set is equinumerous to its converse. (Contributed by Mario Carneiro, 28-Dec-2014.)
((Rel 𝐴𝐴𝑉) → 𝐴𝐴)
 
Theoremcnvct 6803 If a set is dominated by ω, so is its converse. (Contributed by Thierry Arnoux, 29-Dec-2016.)
(𝐴 ≼ ω → 𝐴 ≼ ω)
 
Theoremfndmeng 6804 A function is equinumerate to its domain. (Contributed by Paul Chapman, 22-Jun-2011.)
((𝐹 Fn 𝐴𝐴𝐶) → 𝐴𝐹)
 
Theoremmapsnen 6805 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 6806 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 6807 Two singletons are equinumerous. (Contributed by NM, 9-Nov-2003.)
((𝐴𝐶𝐵𝐷) → {𝐴} ≈ {𝐵})
 
Theoremsnfig 6808 A singleton is finite. For the proper class case, see snprc 3656. (Contributed by Jim Kingdon, 13-Apr-2020.)
(𝐴𝑉 → {𝐴} ∈ Fin)
 
Theoremfiprc 6809 The class of finite sets is a proper class. (Contributed by Jeff Hankins, 3-Oct-2008.)
Fin ∉ V
 
Theoremunen 6810 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 6811 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 6812 A subset of a set dominated by ω is dominated by ω. (Contributed by Thierry Arnoux, 31-Jan-2017.)
((𝐴𝐵𝐵 ≼ ω) → 𝐴 ≼ ω)
 
Theorem1domsn 6813 A singleton (whether of a set or a proper class) is dominated by one. (Contributed by Jim Kingdon, 1-Mar-2022.)
{𝐴} ≼ 1o
 
Theoremenm 6814* A set equinumerous to an inhabited set is inhabited. (Contributed by Jim Kingdon, 19-May-2020.)
((𝐴𝐵 ∧ ∃𝑥 𝑥𝐴) → ∃𝑦 𝑦𝐵)
 
Theoremxpsnen 6815 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 6816 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 6817 One times a cardinal number. (Contributed by NM, 27-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
(𝐴𝑉 → (𝐴 × 1o) ≈ 𝐴)
 
Theoremendisj 6818* 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 6819* 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 6820* Composition with the bijection of xpcomf1o 6819 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 6821 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 6822 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 6823 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 6824 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 6825 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 6826 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 6827 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 6828* 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 6829 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))
 
Theoremfopwdom 6830 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 6831 Any set dominates the empty set. (Contributed by NM, 26-Oct-2003.) (Revised by Mario Carneiro, 26-Apr-2015.)
(𝐴𝑉 → ∅ ≼ 𝐴)
 
Theoremdom0 6832 A set dominated by the empty set is empty. (Contributed by NM, 22-Nov-2004.)
(𝐴 ≼ ∅ ↔ 𝐴 = ∅)
 
Theorem0dom 6833 Any set dominates the empty set. (Contributed by NM, 26-Oct-2003.) (Revised by Mario Carneiro, 26-Apr-2015.)
𝐴 ∈ V       ∅ ≼ 𝐴
 
Theoremenen1 6834 Equality-like theorem for equinumerosity. (Contributed by NM, 18-Dec-2003.)
(𝐴𝐵 → (𝐴𝐶𝐵𝐶))
 
Theoremenen2 6835 Equality-like theorem for equinumerosity. (Contributed by NM, 18-Dec-2003.)
(𝐴𝐵 → (𝐶𝐴𝐶𝐵))
 
Theoremdomen1 6836 Equality-like theorem for equinumerosity and dominance. (Contributed by NM, 8-Nov-2003.)
(𝐴𝐵 → (𝐴𝐶𝐵𝐶))
 
Theoremdomen2 6837 Equality-like theorem for equinumerosity and dominance. (Contributed by NM, 8-Nov-2003.)
(𝐴𝐵 → (𝐶𝐴𝐶𝐵))
 
2.6.29  Equinumerosity (cont.)
 
Theoremxpf1o 6838* 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 6839 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 6840 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 6841 Order-preserving property of set exponentiation. (Contributed by Jim Kingdon, 15-Jul-2022.)
((𝐴𝐵𝐶𝑉) → (𝐴𝑚 𝐶) ≼ (𝐵𝑚 𝐶))
 
Theoremmapxpen 6842 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 6843* Lemma for xpmapen 6844. (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 6844 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 6845* 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 6846 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 6847 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 6848 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 6850. (Contributed by NM, 26-May-1998.)
𝐴 ∈ V    &   𝐵 ∈ V       ((𝐴 ∈ ω ∧ 𝐵 ∈ suc 𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵}))
 
Theoremphplem4 6849 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 6850 A natural number is equinumerous to its successor minus any element of the successor. Version of phplem3 6848 with unnecessary hypotheses removed. (Contributed by Jim Kingdon, 1-Sep-2021.)
((𝐴 ∈ ω ∧ 𝐵 ∈ suc 𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵}))
 
Theoremnneneq 6851 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 6852 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 6853 A singleton {𝐴} is never equinumerous with the ordinal number 2. If 𝐴 is a proper class, see snnen2oprc 6854. (Contributed by Jim Kingdon, 1-Sep-2021.)
(𝐴𝑉 → ¬ {𝐴} ≈ 2o)
 
Theoremsnnen2oprc 6854 A singleton {𝐴} is never equinumerous with the ordinal number 2. If 𝐴 is a set, see snnen2og 6853. (Contributed by Jim Kingdon, 1-Sep-2021.)
𝐴 ∈ V → ¬ {𝐴} ≈ 2o)
 
Theorem1nen2 6855 One and two are not equinumerous. (Contributed by Jim Kingdon, 25-Jan-2022.)
¬ 1o ≈ 2o
 
Theoremphplem4dom 6856 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 6857 A natural number does not dominate its successor. (Contributed by Jim Kingdon, 1-Sep-2021.)
(𝐴 ∈ ω → ¬ suc 𝐴𝐴)
 
Theoremnndomo 6858 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 6859* 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 6846 through phplem4 6849, nneneq 6851, and this final piece of the proof. (Contributed by NM, 29-May-1998.)
((𝐴 ∈ ω ∧ 𝐵𝐴 ∧ ∃𝑥 𝑥 ∈ (𝐴𝐵)) → ¬ 𝐴𝐵)
 
Theoremphpelm 6860 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 6861 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 6862 A finite set is dominated by ω. Also see finct 7109. (Contributed by Thierry Arnoux, 27-Mar-2018.)
(𝐴 ∈ Fin → 𝐴 ≼ ω)
 
Theoremfidceq 6863 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 6864 All decrements of a finite set are equinumerous. (Contributed by Jim Kingdon, 9-Sep-2021.)
((𝑋 ∈ Fin ∧ 𝐴𝑋𝐵𝑋) → (𝑋 ∖ {𝐴}) ≈ (𝑋 ∖ {𝐵}))
 
Theoremfidifsnid 6865 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 3737 from subset to equality when the set is finite. (Contributed by Jim Kingdon, 9-Sep-2021.)
((𝐴 ∈ Fin ∧ 𝐵𝐴) → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) = 𝐴)
 
Theoremnnfi 6866 Natural numbers are finite sets. (Contributed by Stefan O'Rear, 21-Mar-2015.)
(𝐴 ∈ ω → 𝐴 ∈ Fin)
 
Theoremenfi 6867 Equinumerous sets have the same finiteness. (Contributed by NM, 22-Aug-2008.)
(𝐴𝐵 → (𝐴 ∈ Fin ↔ 𝐵 ∈ Fin))
 
Theoremenfii 6868 A set equinumerous to a finite set is finite. (Contributed by Mario Carneiro, 12-Mar-2015.)
((𝐵 ∈ Fin ∧ 𝐴𝐵) → 𝐴 ∈ Fin)
 
Theoremssfilem 6869* Lemma for ssfiexmid 6870. (Contributed by Jim Kingdon, 3-Feb-2022.)
{𝑧 ∈ {∅} ∣ 𝜑} ∈ Fin       (𝜑 ∨ ¬ 𝜑)
 
Theoremssfiexmid 6870* 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.)
𝑥𝑦((𝑥 ∈ Fin ∧ 𝑦𝑥) → 𝑦 ∈ Fin)       (𝜑 ∨ ¬ 𝜑)
 
Theoreminfiexmid 6871* 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 → (𝑥𝑦) ∈ Fin)       (𝜑 ∨ ¬ 𝜑)
 
Theoremdomfiexmid 6872* If any set dominated by a finite set is finite, excluded middle follows. (Contributed by Jim Kingdon, 3-Feb-2022.)
((𝑥 ∈ Fin ∧ 𝑦𝑥) → 𝑦 ∈ Fin)       (𝜑 ∨ ¬ 𝜑)
 
Theoremdif1en 6873 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 𝑀𝑋𝐴) → (𝐴 ∖ {𝑋}) ≈ 𝑀)
 
Theoremdif1enen 6874 Subtracting one element from each of two equinumerous finite sets. (Contributed by Jim Kingdon, 5-Jun-2022.)
(𝜑𝐴 ∈ Fin)    &   (𝜑𝐴𝐵)    &   (𝜑𝐶𝐴)    &   (𝜑𝐷𝐵)       (𝜑 → (𝐴 ∖ {𝐶}) ≈ (𝐵 ∖ {𝐷}))
 
Theoremfiunsnnn 6875 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 ∧ 𝐴𝑁)) → (𝐴 ∪ {𝐵}) ≈ suc 𝑁)
 
Theoremphp5fin 6876 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}))
 
Theoremfisbth 6877 Schroeder-Bernstein Theorem for finite sets. (Contributed by Jim Kingdon, 12-Sep-2021.)
(((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) ∧ (𝐴𝐵𝐵𝐴)) → 𝐴𝐵)
 
Theorem0fin 6878 The empty set is finite. (Contributed by FL, 14-Jul-2008.)
∅ ∈ Fin
 
Theoremfin0 6879* A nonempty finite set has at least one element. (Contributed by Jim Kingdon, 10-Sep-2021.)
(𝐴 ∈ Fin → (𝐴 ≠ ∅ ↔ ∃𝑥 𝑥𝐴))
 
Theoremfin0or 6880* A finite set is either empty or inhabited. (Contributed by Jim Kingdon, 30-Sep-2021.)
(𝐴 ∈ Fin → (𝐴 = ∅ ∨ ∃𝑥 𝑥𝐴))
 
Theoremdiffitest 6881* 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 → (𝐴𝐵) ∈ Fin. (Contributed by Jim Kingdon, 8-Sep-2021.)
𝑎 ∈ Fin ∀𝑏(𝑎𝑏) ∈ Fin       𝜑 ∨ ¬ ¬ 𝜑)
 
Theoremfindcard 6882* 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 = ∅ → (𝜑𝜓))    &   (𝑥 = (𝑦 ∖ {𝑧}) → (𝜑𝜒))    &   (𝑥 = 𝑦 → (𝜑𝜃))    &   (𝑥 = 𝐴 → (𝜑𝜏))    &   𝜓    &   (𝑦 ∈ Fin → (∀𝑧𝑦 𝜒𝜃))       (𝐴 ∈ Fin → 𝜏)
 
Theoremfindcard2 6883* 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 = ∅ → (𝜑𝜓))    &   (𝑥 = 𝑦 → (𝜑𝜒))    &   (𝑥 = (𝑦 ∪ {𝑧}) → (𝜑𝜃))    &   (𝑥 = 𝐴 → (𝜑𝜏))    &   𝜓    &   (𝑦 ∈ Fin → (𝜒𝜃))       (𝐴 ∈ Fin → 𝜏)
 
Theoremfindcard2s 6884* Variation of findcard2 6883 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 = ∅ → (𝜑𝜓))    &   (𝑥 = 𝑦 → (𝜑𝜒))    &   (𝑥 = (𝑦 ∪ {𝑧}) → (𝜑𝜃))    &   (𝑥 = 𝐴 → (𝜑𝜏))    &   𝜓    &   ((𝑦 ∈ Fin ∧ ¬ 𝑧𝑦) → (𝜒𝜃))       (𝐴 ∈ Fin → 𝜏)
 
Theoremfindcard2d 6885* Deduction version of findcard2 6883. If you also need 𝑦 ∈ Fin (which doesn't come for free due to ssfiexmid 6870), use findcard2sd 6886 instead. (Contributed by SO, 16-Jul-2018.)
(𝑥 = ∅ → (𝜓𝜒))    &   (𝑥 = 𝑦 → (𝜓𝜃))    &   (𝑥 = (𝑦 ∪ {𝑧}) → (𝜓𝜏))    &   (𝑥 = 𝐴 → (𝜓𝜂))    &   (𝜑𝜒)    &   ((𝜑 ∧ (𝑦𝐴𝑧 ∈ (𝐴𝑦))) → (𝜃𝜏))    &   (𝜑𝐴 ∈ Fin)       (𝜑𝜂)
 
Theoremfindcard2sd 6886* Deduction form of finite set induction . (Contributed by Jim Kingdon, 14-Sep-2021.)
(𝑥 = ∅ → (𝜓𝜒))    &   (𝑥 = 𝑦 → (𝜓𝜃))    &   (𝑥 = (𝑦 ∪ {𝑧}) → (𝜓𝜏))    &   (𝑥 = 𝐴 → (𝜓𝜂))    &   (𝜑𝜒)    &   (((𝜑𝑦 ∈ Fin) ∧ (𝑦𝐴𝑧 ∈ (𝐴𝑦))) → (𝜃𝜏))    &   (𝜑𝐴 ∈ Fin)       (𝜑𝜂)
 
Theoremdiffisn 6887 Subtracting a singleton from a finite set produces a finite set. (Contributed by Jim Kingdon, 11-Sep-2021.)
((𝐴 ∈ Fin ∧ 𝐵𝐴) → (𝐴 ∖ {𝐵}) ∈ Fin)
 
Theoremdiffifi 6888 Subtracting one finite set from another produces a finite set. (Contributed by Jim Kingdon, 8-Sep-2021.)
((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin ∧ 𝐵𝐴) → (𝐴𝐵) ∈ Fin)
 
Theoreminfnfi 6889 An infinite set is not finite. (Contributed by Jim Kingdon, 20-Feb-2022.)
(ω ≼ 𝐴 → ¬ 𝐴 ∈ Fin)
 
Theoremominf 6890 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 6761. (Contributed by NM, 2-Jun-1998.)
¬ ω ∈ Fin
 
Theoremisinfinf 6891* An infinite set contains subsets of arbitrarily large finite cardinality. (Contributed by Jim Kingdon, 15-Jun-2022.)
(ω ≼ 𝐴 → ∀𝑛 ∈ ω ∃𝑥(𝑥𝐴𝑥𝑛))
 
Theoremac6sfi 6892* Existence of a choice function for finite sets. (Contributed by Jeff Hankins, 26-Jun-2009.) (Proof shortened by Mario Carneiro, 29-Jan-2014.)
(𝑦 = (𝑓𝑥) → (𝜑𝜓))       ((𝐴 ∈ Fin ∧ ∀𝑥𝐴𝑦𝐵 𝜑) → ∃𝑓(𝑓:𝐴𝐵 ∧ ∀𝑥𝐴 𝜓))
 
Theoremtridc 6893* A trichotomous order is decidable. (Contributed by Jim Kingdon, 5-Sep-2022.)
(𝜑𝑅 Po 𝐴)    &   (𝜑 → ∀𝑥𝐴𝑦𝐴 (𝑥𝑅𝑦𝑥 = 𝑦𝑦𝑅𝑥))    &   (𝜑𝐵𝐴)    &   (𝜑𝐶𝐴)       (𝜑DECID 𝐵𝑅𝐶)
 
Theoremfimax2gtrilemstep 6894* Lemma for fimax2gtri 6895. The induction step. (Contributed by Jim Kingdon, 5-Sep-2022.)
(𝜑𝑅 Po 𝐴)    &   (𝜑 → ∀𝑥𝐴𝑦𝐴 (𝑥𝑅𝑦𝑥 = 𝑦𝑦𝑅𝑥))    &   (𝜑𝐴 ∈ Fin)    &   (𝜑𝐴 ≠ ∅)    &   (𝜑𝑈 ∈ Fin)    &   (𝜑𝑈𝐴)    &   (𝜑𝑍𝐴)    &   (𝜑𝑉𝐴)    &   (𝜑 → ¬ 𝑉𝑈)    &   (𝜑 → ∀𝑦𝑈 ¬ 𝑍𝑅𝑦)       (𝜑 → ∃𝑥𝐴𝑦 ∈ (𝑈 ∪ {𝑉}) ¬ 𝑥𝑅𝑦)
 
Theoremfimax2gtri 6895* A finite set has a maximum under a trichotomous order. (Contributed by Jim Kingdon, 5-Sep-2022.)
(𝜑𝑅 Po 𝐴)    &   (𝜑 → ∀𝑥𝐴𝑦𝐴 (𝑥𝑅𝑦𝑥 = 𝑦𝑦𝑅𝑥))    &   (𝜑𝐴 ∈ Fin)    &   (𝜑𝐴 ≠ ∅)       (𝜑 → ∃𝑥𝐴𝑦𝐴 ¬ 𝑥𝑅𝑦)
 
Theoremfinexdc 6896* Decidability of existence, over a finite set and defined by a decidable proposition. (Contributed by Jim Kingdon, 12-Jul-2022.)
((𝐴 ∈ Fin ∧ ∀𝑥𝐴 DECID 𝜑) → DECID𝑥𝐴 𝜑)
 
Theoremdfrex2fin 6897* Relationship between universal and existential quantifiers over a finite set. Remark in Section 2.2.1 of [Pierik], p. 8. Although Pierik does not mention the decidability condition explicitly, it does say "only finitely many x to check" which means there must be some way of checking each value of x. (Contributed by Jim Kingdon, 11-Jul-2022.)
((𝐴 ∈ Fin ∧ ∀𝑥𝐴 DECID 𝜑) → (∃𝑥𝐴 𝜑 ↔ ¬ ∀𝑥𝐴 ¬ 𝜑))
 
Theoreminfm 6898* An infinite set is inhabited. (Contributed by Jim Kingdon, 18-Feb-2022.)
(ω ≼ 𝐴 → ∃𝑥 𝑥𝐴)
 
Theoreminfn0 6899 An infinite set is not empty. (Contributed by NM, 23-Oct-2004.)
(ω ≼ 𝐴𝐴 ≠ ∅)
 
Theoreminffiexmid 6900* If any given set is either finite or infinite, excluded middle follows. (Contributed by Jim Kingdon, 15-Jun-2022.)
(𝑥 ∈ Fin ∨ ω ≼ 𝑥)       (𝜑 ∨ ¬ 𝜑)
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