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Theorem List for Intuitionistic Logic Explorer - 6701-6800   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremen1bg 6701 A set is equinumerous to ordinal one iff it is a singleton. (Contributed by Jim Kingdon, 13-Apr-2020.)
(𝐴𝑉 → (𝐴 ≈ 1o𝐴 = { 𝐴}))
 
Theoremreuen1 6702* Two ways to express "exactly one". (Contributed by Stefan O'Rear, 28-Oct-2014.)
(∃!𝑥𝐴 𝜑 ↔ {𝑥𝐴𝜑} ≈ 1o)
 
Theoremeuen1 6703 Two ways to express "exactly one". (Contributed by Stefan O'Rear, 28-Oct-2014.)
(∃!𝑥𝜑 ↔ {𝑥𝜑} ≈ 1o)
 
Theoremeuen1b 6704* Two ways to express "𝐴 has a unique element". (Contributed by Mario Carneiro, 9-Apr-2015.)
(𝐴 ≈ 1o ↔ ∃!𝑥 𝑥𝐴)
 
Theoremen1uniel 6705 A singleton contains its sole element. (Contributed by Stefan O'Rear, 16-Aug-2015.)
(𝑆 ≈ 1o 𝑆𝑆)
 
Theorem2dom 6706* A set that dominates ordinal 2 has at least 2 different members. (Contributed by NM, 25-Jul-2004.)
(2o𝐴 → ∃𝑥𝐴𝑦𝐴 ¬ 𝑥 = 𝑦)
 
Theoremfundmen 6707 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 6708 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 6709 A relational set is equinumerous to its converse. (Contributed by Mario Carneiro, 28-Dec-2014.)
((Rel 𝐴𝐴𝑉) → 𝐴𝐴)
 
Theoremcnvct 6710 If a set is dominated by ω, so is its converse. (Contributed by Thierry Arnoux, 29-Dec-2016.)
(𝐴 ≼ ω → 𝐴 ≼ ω)
 
Theoremfndmeng 6711 A function is equinumerate to its domain. (Contributed by Paul Chapman, 22-Jun-2011.)
((𝐹 Fn 𝐴𝐴𝐶) → 𝐴𝐹)
 
Theoremmapsnen 6712 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 6713 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 6714 Two singletons are equinumerous. (Contributed by NM, 9-Nov-2003.)
((𝐴𝐶𝐵𝐷) → {𝐴} ≈ {𝐵})
 
Theoremsnfig 6715 A singleton is finite. For the proper class case, see snprc 3595. (Contributed by Jim Kingdon, 13-Apr-2020.)
(𝐴𝑉 → {𝐴} ∈ Fin)
 
Theoremfiprc 6716 The class of finite sets is a proper class. (Contributed by Jeff Hankins, 3-Oct-2008.)
Fin ∉ V
 
Theoremunen 6717 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 6718 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 6719 A subset of a set dominated by ω is dominated by ω. (Contributed by Thierry Arnoux, 31-Jan-2017.)
((𝐴𝐵𝐵 ≼ ω) → 𝐴 ≼ ω)
 
Theorem1domsn 6720 A singleton (whether of a set or a proper class) is dominated by one. (Contributed by Jim Kingdon, 1-Mar-2022.)
{𝐴} ≼ 1o
 
Theoremenm 6721* A set equinumerous to an inhabited set is inhabited. (Contributed by Jim Kingdon, 19-May-2020.)
((𝐴𝐵 ∧ ∃𝑥 𝑥𝐴) → ∃𝑦 𝑦𝐵)
 
Theoremxpsnen 6722 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 6723 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 6724 One times a cardinal number. (Contributed by NM, 27-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
(𝐴𝑉 → (𝐴 × 1o) ≈ 𝐴)
 
Theoremendisj 6725* 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 6726* 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 6727* Composition with the bijection of xpcomf1o 6726 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 6728 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 6729 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 6730 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 6731 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 6732 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 6733 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 6734 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 6735* 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 6736 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 6737 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 6738 Any set dominates the empty set. (Contributed by NM, 26-Oct-2003.) (Revised by Mario Carneiro, 26-Apr-2015.)
(𝐴𝑉 → ∅ ≼ 𝐴)
 
Theoremdom0 6739 A set dominated by the empty set is empty. (Contributed by NM, 22-Nov-2004.)
(𝐴 ≼ ∅ ↔ 𝐴 = ∅)
 
Theorem0dom 6740 Any set dominates the empty set. (Contributed by NM, 26-Oct-2003.) (Revised by Mario Carneiro, 26-Apr-2015.)
𝐴 ∈ V       ∅ ≼ 𝐴
 
Theoremenen1 6741 Equality-like theorem for equinumerosity. (Contributed by NM, 18-Dec-2003.)
(𝐴𝐵 → (𝐴𝐶𝐵𝐶))
 
Theoremenen2 6742 Equality-like theorem for equinumerosity. (Contributed by NM, 18-Dec-2003.)
(𝐴𝐵 → (𝐶𝐴𝐶𝐵))
 
Theoremdomen1 6743 Equality-like theorem for equinumerosity and dominance. (Contributed by NM, 8-Nov-2003.)
(𝐴𝐵 → (𝐴𝐶𝐵𝐶))
 
Theoremdomen2 6744 Equality-like theorem for equinumerosity and dominance. (Contributed by NM, 8-Nov-2003.)
(𝐴𝐵 → (𝐶𝐴𝐶𝐵))
 
2.6.28  Equinumerosity (cont.)
 
Theoremxpf1o 6745* 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 6746 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 6747 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 6748 Order-preserving property of set exponentiation. (Contributed by Jim Kingdon, 15-Jul-2022.)
((𝐴𝐵𝐶𝑉) → (𝐴𝑚 𝐶) ≼ (𝐵𝑚 𝐶))
 
Theoremmapxpen 6749 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 6750* Lemma for xpmapen 6751. (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 6751 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 6752* 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)})
 
2.6.29  Pigeonhole Principle
 
Theoremphplem1 6753 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 6754 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 6755 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 6757. (Contributed by NM, 26-May-1998.)
𝐴 ∈ V    &   𝐵 ∈ V       ((𝐴 ∈ ω ∧ 𝐵 ∈ suc 𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵}))
 
Theoremphplem4 6756 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 6757 A natural number is equinumerous to its successor minus any element of the successor. Version of phplem3 6755 with unnecessary hypotheses removed. (Contributed by Jim Kingdon, 1-Sep-2021.)
((𝐴 ∈ ω ∧ 𝐵 ∈ suc 𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵}))
 
Theoremnneneq 6758 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 6759 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 6760 A singleton {𝐴} is never equinumerous with the ordinal number 2. If 𝐴 is a proper class, see snnen2oprc 6761. (Contributed by Jim Kingdon, 1-Sep-2021.)
(𝐴𝑉 → ¬ {𝐴} ≈ 2o)
 
Theoremsnnen2oprc 6761 A singleton {𝐴} is never equinumerous with the ordinal number 2. If 𝐴 is a set, see snnen2og 6760. (Contributed by Jim Kingdon, 1-Sep-2021.)
𝐴 ∈ V → ¬ {𝐴} ≈ 2o)
 
Theorem1nen2 6762 One and two are not equinumerous. (Contributed by Jim Kingdon, 25-Jan-2022.)
¬ 1o ≈ 2o
 
Theoremphplem4dom 6763 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 6764 A natural number does not dominate its successor. (Contributed by Jim Kingdon, 1-Sep-2021.)
(𝐴 ∈ ω → ¬ suc 𝐴𝐴)
 
Theoremnndomo 6765 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 6766* 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 6753 through phplem4 6756, nneneq 6758, and this final piece of the proof. (Contributed by NM, 29-May-1998.)
((𝐴 ∈ ω ∧ 𝐵𝐴 ∧ ∃𝑥 𝑥 ∈ (𝐴𝐵)) → ¬ 𝐴𝐵)
 
Theoremphpelm 6767 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 6768 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))
 
2.6.30  Finite sets
 
Theoremfict 6769 A finite set is dominated by ω. Also see finct 7008. (Contributed by Thierry Arnoux, 27-Mar-2018.)
(𝐴 ∈ Fin → 𝐴 ≼ ω)
 
Theoremfidceq 6770 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 6771 All decrements of a finite set are equinumerous. (Contributed by Jim Kingdon, 9-Sep-2021.)
((𝑋 ∈ Fin ∧ 𝐴𝑋𝐵𝑋) → (𝑋 ∖ {𝐴}) ≈ (𝑋 ∖ {𝐵}))
 
Theoremfidifsnid 6772 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 3673 from subset to equality when the set is finite. (Contributed by Jim Kingdon, 9-Sep-2021.)
((𝐴 ∈ Fin ∧ 𝐵𝐴) → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) = 𝐴)
 
Theoremnnfi 6773 Natural numbers are finite sets. (Contributed by Stefan O'Rear, 21-Mar-2015.)
(𝐴 ∈ ω → 𝐴 ∈ Fin)
 
Theoremenfi 6774 Equinumerous sets have the same finiteness. (Contributed by NM, 22-Aug-2008.)
(𝐴𝐵 → (𝐴 ∈ Fin ↔ 𝐵 ∈ Fin))
 
Theoremenfii 6775 A set equinumerous to a finite set is finite. (Contributed by Mario Carneiro, 12-Mar-2015.)
((𝐵 ∈ Fin ∧ 𝐴𝐵) → 𝐴 ∈ Fin)
 
Theoremssfilem 6776* Lemma for ssfiexmid 6777. (Contributed by Jim Kingdon, 3-Feb-2022.)
{𝑧 ∈ {∅} ∣ 𝜑} ∈ Fin       (𝜑 ∨ ¬ 𝜑)
 
Theoremssfiexmid 6777* 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 6778* 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 6779* 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 6780 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 6781 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 6782 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 6783 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 6784 Schroeder-Bernstein Theorem for finite sets. (Contributed by Jim Kingdon, 12-Sep-2021.)
(((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) ∧ (𝐴𝐵𝐵𝐴)) → 𝐴𝐵)
 
Theorem0fin 6785 The empty set is finite. (Contributed by FL, 14-Jul-2008.)
∅ ∈ Fin
 
Theoremfin0 6786* A nonempty finite set has at least one element. (Contributed by Jim Kingdon, 10-Sep-2021.)
(𝐴 ∈ Fin → (𝐴 ≠ ∅ ↔ ∃𝑥 𝑥𝐴))
 
Theoremfin0or 6787* A finite set is either empty or inhabited. (Contributed by Jim Kingdon, 30-Sep-2021.)
(𝐴 ∈ Fin → (𝐴 = ∅ ∨ ∃𝑥 𝑥𝐴))
 
Theoremdiffitest 6788* 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 6789* 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 6790* 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 6791* Variation of findcard2 6790 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 6792* Deduction version of findcard2 6790. If you also need 𝑦 ∈ Fin (which doesn't come for free due to ssfiexmid 6777), use findcard2sd 6793 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 6793* 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 6794 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 6795 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 6796 An infinite set is not finite. (Contributed by Jim Kingdon, 20-Feb-2022.)
(ω ≼ 𝐴 → ¬ 𝐴 ∈ Fin)
 
Theoremominf 6797 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 6668. (Contributed by NM, 2-Jun-1998.)
¬ ω ∈ Fin
 
Theoremisinfinf 6798* An infinite set contains subsets of arbitrarily large finite cardinality. (Contributed by Jim Kingdon, 15-Jun-2022.)
(ω ≼ 𝐴 → ∀𝑛 ∈ ω ∃𝑥(𝑥𝐴𝑥𝑛))
 
Theoremac6sfi 6799* 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 6800* A trichotomous order is decidable. (Contributed by Jim Kingdon, 5-Sep-2022.)
(𝜑𝑅 Po 𝐴)    &   (𝜑 → ∀𝑥𝐴𝑦𝐴 (𝑥𝑅𝑦𝑥 = 𝑦𝑦𝑅𝑥))    &   (𝜑𝐵𝐴)    &   (𝜑𝐶𝐴)       (𝜑DECID 𝐵𝑅𝐶)
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