Theorem List for Intuitionistic Logic Explorer - 6901-7000 *Has distinct variable
group(s)
| Type | Label | Description |
| Statement |
| |
| Theorem | ener 6901 |
Equinumerosity is an equivalence relation. (Contributed by NM,
19-Mar-1998.) (Revised by Mario Carneiro, 15-Nov-2014.)
|
| ⊢ ≈ Er V |
| |
| Theorem | ensymb 6902 |
Symmetry of equinumerosity. Theorem 2 of [Suppes] p. 92. (Contributed by
Mario Carneiro, 26-Apr-2015.)
|
| ⊢ (𝐴 ≈ 𝐵 ↔ 𝐵 ≈ 𝐴) |
| |
| Theorem | ensym 6903 |
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 6904 |
Symmetry of equinumerosity. Theorem 2 of [Suppes] p. 92. (Contributed
by NM, 25-Sep-2004.)
|
| ⊢ 𝐴 ≈ 𝐵 ⇒ ⊢ 𝐵 ≈ 𝐴 |
| |
| Theorem | ensymd 6905 |
Symmetry of equinumerosity. Deduction form of ensym 6903. (Contributed
by David Moews, 1-May-2017.)
|
| ⊢ (𝜑 → 𝐴 ≈ 𝐵) ⇒ ⊢ (𝜑 → 𝐵 ≈ 𝐴) |
| |
| Theorem | entr 6906 |
Transitivity of equinumerosity. Theorem 3 of [Suppes] p. 92.
(Contributed by NM, 9-Jun-1998.)
|
| ⊢ ((𝐴 ≈ 𝐵 ∧ 𝐵 ≈ 𝐶) → 𝐴 ≈ 𝐶) |
| |
| Theorem | domtr 6907 |
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 6908 |
A chained equinumerosity inference. (Contributed by NM,
25-Sep-2004.)
|
| ⊢ 𝐴 ≈ 𝐵
& ⊢ 𝐵 ≈ 𝐶 ⇒ ⊢ 𝐴 ≈ 𝐶 |
| |
| Theorem | entr2i 6909 |
A chained equinumerosity inference. (Contributed by NM,
25-Sep-2004.)
|
| ⊢ 𝐴 ≈ 𝐵
& ⊢ 𝐵 ≈ 𝐶 ⇒ ⊢ 𝐶 ≈ 𝐴 |
| |
| Theorem | entr3i 6910 |
A chained equinumerosity inference. (Contributed by NM,
25-Sep-2004.)
|
| ⊢ 𝐴 ≈ 𝐵
& ⊢ 𝐴 ≈ 𝐶 ⇒ ⊢ 𝐵 ≈ 𝐶 |
| |
| Theorem | entr4i 6911 |
A chained equinumerosity inference. (Contributed by NM,
25-Sep-2004.)
|
| ⊢ 𝐴 ≈ 𝐵
& ⊢ 𝐶 ≈ 𝐵 ⇒ ⊢ 𝐴 ≈ 𝐶 |
| |
| Theorem | endomtr 6912 |
Transitivity of equinumerosity and dominance. (Contributed by NM,
7-Jun-1998.)
|
| ⊢ ((𝐴 ≈ 𝐵 ∧ 𝐵 ≼ 𝐶) → 𝐴 ≼ 𝐶) |
| |
| Theorem | domentr 6913 |
Transitivity of dominance and equinumerosity. (Contributed by NM,
7-Jun-1998.)
|
| ⊢ ((𝐴 ≼ 𝐵 ∧ 𝐵 ≈ 𝐶) → 𝐴 ≼ 𝐶) |
| |
| Theorem | f1imaeng 6914 |
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 6915 |
A one-to-one function's image under a subset of its domain is equinumerous
to the subset. (This version of f1imaen 6916 does not need ax-setind 4606.)
(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 6916 |
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 6917 |
The empty set is equinumerous only to itself. Exercise 1 of
[TakeutiZaring] p. 88.
(Contributed by NM, 27-May-1998.)
|
| ⊢ (𝐴 ≈ ∅ ↔ 𝐴 = ∅) |
| |
| Theorem | ensn1 6918 |
A singleton is equinumerous to ordinal one. (Contributed by NM,
4-Nov-2002.)
|
| ⊢ 𝐴 ∈ V ⇒ ⊢ {𝐴} ≈ 1o |
| |
| Theorem | ensn1g 6919 |
A singleton is equinumerous to ordinal one. (Contributed by NM,
23-Apr-2004.)
|
| ⊢ (𝐴 ∈ 𝑉 → {𝐴} ≈ 1o) |
| |
| Theorem | enpr1g 6920 |
{𝐴, 𝐴} has only one element.
(Contributed by FL, 15-Feb-2010.)
|
| ⊢ (𝐴 ∈ 𝑉 → {𝐴, 𝐴} ≈ 1o) |
| |
| Theorem | en1 6921* |
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 6922 |
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 6923* |
Two ways to express "exactly one". (Contributed by Stefan O'Rear,
28-Oct-2014.)
|
| ⊢ (∃!𝑥 ∈ 𝐴 𝜑 ↔ {𝑥 ∈ 𝐴 ∣ 𝜑} ≈ 1o) |
| |
| Theorem | euen1 6924 |
Two ways to express "exactly one". (Contributed by Stefan O'Rear,
28-Oct-2014.)
|
| ⊢ (∃!𝑥𝜑 ↔ {𝑥 ∣ 𝜑} ≈ 1o) |
| |
| Theorem | euen1b 6925* |
Two ways to express "𝐴 has a unique element".
(Contributed by
Mario Carneiro, 9-Apr-2015.)
|
| ⊢ (𝐴 ≈ 1o ↔ ∃!𝑥 𝑥 ∈ 𝐴) |
| |
| Theorem | en1uniel 6926 |
A singleton contains its sole element. (Contributed by Stefan O'Rear,
16-Aug-2015.)
|
| ⊢ (𝑆 ≈ 1o → ∪ 𝑆
∈ 𝑆) |
| |
| Theorem | en1m 6927* |
A set with one element is inhabited. (Contributed by Jim Kingdon,
3-Jan-2026.)
|
| ⊢ (𝐴 ≈ 1o → ∃𝑥 𝑥 ∈ 𝐴) |
| |
| Theorem | 2dom 6928* |
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 6929 |
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 6930 |
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 6931 |
A relational set is equinumerous to its converse. (Contributed by Mario
Carneiro, 28-Dec-2014.)
|
| ⊢ ((Rel 𝐴 ∧ 𝐴 ∈ 𝑉) → 𝐴 ≈ ◡𝐴) |
| |
| Theorem | cnvct 6932 |
If a set is dominated by ω, so is its converse.
(Contributed by
Thierry Arnoux, 29-Dec-2016.)
|
| ⊢ (𝐴 ≼ ω → ◡𝐴 ≼ ω) |
| |
| Theorem | fndmeng 6933 |
A function is equinumerate to its domain. (Contributed by Paul Chapman,
22-Jun-2011.)
|
| ⊢ ((𝐹 Fn 𝐴 ∧ 𝐴 ∈ 𝐶) → 𝐴 ≈ 𝐹) |
| |
| Theorem | mapsnen 6934 |
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 6935 |
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 6936 |
Two singletons are equinumerous. (Contributed by NM, 9-Nov-2003.)
|
| ⊢ ((𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷) → {𝐴} ≈ {𝐵}) |
| |
| Theorem | snfig 6937 |
A singleton is finite. For the proper class case, see snprc 3711.
(Contributed by Jim Kingdon, 13-Apr-2020.)
|
| ⊢ (𝐴 ∈ 𝑉 → {𝐴} ∈ Fin) |
| |
| Theorem | fiprc 6938 |
The class of finite sets is a proper class. (Contributed by Jeff
Hankins, 3-Oct-2008.)
|
| ⊢ Fin ∉ V |
| |
| Theorem | unen 6939 |
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 6940 |
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 6941* |
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 6942 |
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 6943* |
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 | en2m 6944* |
A set with two elements is inhabited. (Contributed by Jim Kingdon,
3-Jan-2026.)
|
| ⊢ (𝐴 ≈ 2o → ∃𝑥 𝑥 ∈ 𝐴) |
| |
| Theorem | ssct 6945 |
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 6946 |
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 6947* |
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 6948 |
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 6949 |
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 6950 |
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 6951* |
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 6952* |
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 6953* |
Composition with the bijection of xpcomf1o 6952 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 6954 |
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 6955 |
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 6956 |
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 6957 |
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 6958 |
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 6959 |
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 6960 |
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 6961* |
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 6962 |
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 6963* |
Lemma for pw2f1odc 6964. (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 6964* |
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 6965 |
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 6966 |
Any set dominates the empty set. (Contributed by NM, 26-Oct-2003.)
(Revised by Mario Carneiro, 26-Apr-2015.)
|
| ⊢ (𝐴 ∈ 𝑉 → ∅ ≼ 𝐴) |
| |
| Theorem | dom0 6967 |
A set dominated by the empty set is empty. (Contributed by NM,
22-Nov-2004.)
|
| ⊢ (𝐴 ≼ ∅ ↔ 𝐴 = ∅) |
| |
| Theorem | 0dom 6968 |
Any set dominates the empty set. (Contributed by NM, 26-Oct-2003.)
(Revised by Mario Carneiro, 26-Apr-2015.)
|
| ⊢ 𝐴 ∈ V ⇒ ⊢ ∅ ≼ 𝐴 |
| |
| Theorem | enen1 6969 |
Equality-like theorem for equinumerosity. (Contributed by NM,
18-Dec-2003.)
|
| ⊢ (𝐴 ≈ 𝐵 → (𝐴 ≈ 𝐶 ↔ 𝐵 ≈ 𝐶)) |
| |
| Theorem | enen2 6970 |
Equality-like theorem for equinumerosity. (Contributed by NM,
18-Dec-2003.)
|
| ⊢ (𝐴 ≈ 𝐵 → (𝐶 ≈ 𝐴 ↔ 𝐶 ≈ 𝐵)) |
| |
| Theorem | domen1 6971 |
Equality-like theorem for equinumerosity and dominance. (Contributed by
NM, 8-Nov-2003.)
|
| ⊢ (𝐴 ≈ 𝐵 → (𝐴 ≼ 𝐶 ↔ 𝐵 ≼ 𝐶)) |
| |
| Theorem | domen2 6972 |
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 6973* |
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 6974 |
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 6975 |
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 6976 |
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 6977 |
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 6978* |
Lemma for xpmapen 6979. (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 6979 |
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 6980* |
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 6981 |
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 6982 |
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 6983 |
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 6985. (Contributed by NM,
26-May-1998.)
|
| ⊢ 𝐴 ∈ V & ⊢ 𝐵 ∈
V ⇒ ⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ suc 𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵})) |
| |
| Theorem | phplem4 6984 |
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 6985 |
A natural number is equinumerous to its successor minus any element of
the successor. Version of phplem3 6983 with unnecessary hypotheses
removed. (Contributed by Jim Kingdon, 1-Sep-2021.)
|
| ⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ suc 𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵})) |
| |
| Theorem | nneneq 6986 |
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 6987 |
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 6988 |
A singleton {𝐴} is never equinumerous with the
ordinal number 2. If
𝐴 is a proper class, see snnen2oprc 6989. (Contributed by Jim Kingdon,
1-Sep-2021.)
|
| ⊢ (𝐴 ∈ 𝑉 → ¬ {𝐴} ≈ 2o) |
| |
| Theorem | snnen2oprc 6989 |
A singleton {𝐴} is never equinumerous with the
ordinal number 2. If
𝐴 is a set, see snnen2og 6988. (Contributed by Jim Kingdon,
1-Sep-2021.)
|
| ⊢ (¬ 𝐴 ∈ V → ¬ {𝐴} ≈ 2o) |
| |
| Theorem | 1nen2 6990 |
One and two are not equinumerous. (Contributed by Jim Kingdon,
25-Jan-2022.)
|
| ⊢ ¬ 1o ≈
2o |
| |
| Theorem | phplem4dom 6991 |
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 6992 |
A natural number does not dominate its successor. (Contributed by Jim
Kingdon, 1-Sep-2021.)
|
| ⊢ (𝐴 ∈ ω → ¬ suc 𝐴 ≼ 𝐴) |
| |
| Theorem | nndomo 6993 |
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 | 1ndom2 6994 |
Two is not dominated by one. (Contributed by Jim Kingdon,
10-Jan-2026.)
|
| ⊢ ¬ 2o ≼
1o |
| |
| Theorem | phpm 6995* |
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 6981 through phplem4 6984, nneneq 6986, 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 6996 |
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 6997 |
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 6998 |
A finite set is dominated by ω. Also see finct 7251. (Contributed
by Thierry Arnoux, 27-Mar-2018.)
|
| ⊢ (𝐴 ∈ Fin → 𝐴 ≼ ω) |
| |
| Theorem | fidceq 6999 |
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 7000 |
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})) |