Theorem List for Intuitionistic Logic Explorer - 6801-6900 *Has distinct variable
group(s)
Type | Label | Description |
Statement |
|
Theorem | fundmeng 6801 |
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 6802 |
A relational set is equinumerous to its converse. (Contributed by Mario
Carneiro, 28-Dec-2014.)
|
⊢ ((Rel 𝐴 ∧ 𝐴 ∈ 𝑉) → 𝐴 ≈ ◡𝐴) |
|
Theorem | cnvct 6803 |
If a set is dominated by ω, so is its converse.
(Contributed by
Thierry Arnoux, 29-Dec-2016.)
|
⊢ (𝐴 ≼ ω → ◡𝐴 ≼ ω) |
|
Theorem | fndmeng 6804 |
A function is equinumerate to its domain. (Contributed by Paul Chapman,
22-Jun-2011.)
|
⊢ ((𝐹 Fn 𝐴 ∧ 𝐴 ∈ 𝐶) → 𝐴 ≈ 𝐹) |
|
Theorem | mapsnen 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 ⇒ ⊢ (𝐴 ↑𝑚 {𝐵}) ≈ 𝐴 |
|
Theorem | map1 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.)
|
⊢ (𝐴 ∈ 𝑉 → (1o
↑𝑚 𝐴) ≈ 1o) |
|
Theorem | en2sn 6807 |
Two singletons are equinumerous. (Contributed by NM, 9-Nov-2003.)
|
⊢ ((𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷) → {𝐴} ≈ {𝐵}) |
|
Theorem | snfig 6808 |
A singleton is finite. For the proper class case, see snprc 3656.
(Contributed by Jim Kingdon, 13-Apr-2020.)
|
⊢ (𝐴 ∈ 𝑉 → {𝐴} ∈ Fin) |
|
Theorem | fiprc 6809 |
The class of finite sets is a proper class. (Contributed by Jeff
Hankins, 3-Oct-2008.)
|
⊢ Fin ∉ V |
|
Theorem | unen 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.)
|
⊢ (((𝐴 ≈ 𝐵 ∧ 𝐶 ≈ 𝐷) ∧ ((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐷) = ∅)) → (𝐴 ∪ 𝐶) ≈ (𝐵 ∪ 𝐷)) |
|
Theorem | enpr2d 6811 |
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 | ssct 6812 |
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 6813 |
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 6814* |
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 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 |
|
Theorem | xpsneng 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.)
|
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (𝐴 × {𝐵}) ≈ 𝐴) |
|
Theorem | xp1en 6817 |
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 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 ⇒ ⊢ ∃𝑥∃𝑦((𝑥 ≈ 𝐴 ∧ 𝑦 ≈ 𝐵) ∧ (𝑥 ∩ 𝑦) = ∅) |
|
Theorem | xpcomf1o 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) |
|
Theorem | xpcomco 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 = (𝑦 ∈ 𝐵, 𝑧 ∈ 𝐴 ↦ 𝐶) ⇒ ⊢ (𝐺 ∘ 𝐹) = (𝑧 ∈ 𝐴, 𝑦 ∈ 𝐵 ↦ 𝐶) |
|
Theorem | xpcomen 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) |
|
Theorem | xpcomeng 6822 |
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 6823 |
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 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)) |
|
Theorem | xpdom2 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 ⇒ ⊢ (𝐴 ≼ 𝐵 → (𝐶 × 𝐴) ≼ (𝐶 × 𝐵)) |
|
Theorem | xpdom2g 6826 |
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 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.)
|
⊢ ((𝐶 ∈ 𝑉 ∧ 𝐴 ≼ 𝐵) → (𝐴 × 𝐶) ≼ (𝐵 × 𝐶)) |
|
Theorem | xpdom3m 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.)
|
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ ∃𝑥 𝑥 ∈ 𝐵) → 𝐴 ≼ (𝐴 × 𝐵)) |
|
Theorem | xpdom1 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 ⇒ ⊢ (𝐴 ≼ 𝐵 → (𝐴 × 𝐶) ≼ (𝐵 × 𝐶)) |
|
Theorem | fopwdom 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:𝐴–onto→𝐵) → 𝒫 𝐵 ≼ 𝒫 𝐴) |
|
Theorem | 0domg 6831 |
Any set dominates the empty set. (Contributed by NM, 26-Oct-2003.)
(Revised by Mario Carneiro, 26-Apr-2015.)
|
⊢ (𝐴 ∈ 𝑉 → ∅ ≼ 𝐴) |
|
Theorem | dom0 6832 |
A set dominated by the empty set is empty. (Contributed by NM,
22-Nov-2004.)
|
⊢ (𝐴 ≼ ∅ ↔ 𝐴 = ∅) |
|
Theorem | 0dom 6833 |
Any set dominates the empty set. (Contributed by NM, 26-Oct-2003.)
(Revised by Mario Carneiro, 26-Apr-2015.)
|
⊢ 𝐴 ∈ V ⇒ ⊢ ∅ ≼ 𝐴 |
|
Theorem | enen1 6834 |
Equality-like theorem for equinumerosity. (Contributed by NM,
18-Dec-2003.)
|
⊢ (𝐴 ≈ 𝐵 → (𝐴 ≈ 𝐶 ↔ 𝐵 ≈ 𝐶)) |
|
Theorem | enen2 6835 |
Equality-like theorem for equinumerosity. (Contributed by NM,
18-Dec-2003.)
|
⊢ (𝐴 ≈ 𝐵 → (𝐶 ≈ 𝐴 ↔ 𝐶 ≈ 𝐵)) |
|
Theorem | domen1 6836 |
Equality-like theorem for equinumerosity and dominance. (Contributed by
NM, 8-Nov-2003.)
|
⊢ (𝐴 ≈ 𝐵 → (𝐴 ≼ 𝐶 ↔ 𝐵 ≼ 𝐶)) |
|
Theorem | domen2 6837 |
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 6838* |
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 6839 |
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 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.)
|
⊢ ((𝐴 ≈ 𝐵 ∧ 𝐶 ≈ 𝐷) → (𝐴 ↑𝑚 𝐶) ≈ (𝐵 ↑𝑚 𝐷)) |
|
Theorem | mapdom1g 6841 |
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 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.)
|
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝐶 ∈ 𝑋) → ((𝐴 ↑𝑚 𝐵) ↑𝑚
𝐶) ≈ (𝐴 ↑𝑚
(𝐵 × 𝐶))) |
|
Theorem | xpmapenlem 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 = (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 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) ≈ ((𝐴 ↑𝑚 𝐶) × (𝐵 ↑𝑚 𝐶)) |
|
Theorem | ssenen 6845* |
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 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 ∧ 𝐵 ∈ 𝐴) → ({𝐴} ∪ (𝐴 ∖ {𝐵})) = (suc 𝐴 ∖ {𝐵})) |
|
Theorem | phplem2 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 ∧ 𝐵 ∈ 𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵})) |
|
Theorem | phplem3 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})) |
|
Theorem | phplem4 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 𝐵 → 𝐴 ≈ 𝐵)) |
|
Theorem | phplem3g 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})) |
|
Theorem | nneneq 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) → (𝐴 ≈ 𝐵 ↔ 𝐴 = 𝐵)) |
|
Theorem | php5 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) |
|
Theorem | snnen2og 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.)
|
⊢ (𝐴 ∈ 𝑉 → ¬ {𝐴} ≈ 2o) |
|
Theorem | snnen2oprc 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) |
|
Theorem | 1nen2 6855 |
One and two are not equinumerous. (Contributed by Jim Kingdon,
25-Jan-2022.)
|
⊢ ¬ 1o ≈
2o |
|
Theorem | phplem4dom 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 𝐵 → 𝐴 ≼ 𝐵)) |
|
Theorem | php5dom 6857 |
A natural number does not dominate its successor. (Contributed by Jim
Kingdon, 1-Sep-2021.)
|
⊢ (𝐴 ∈ ω → ¬ suc 𝐴 ≼ 𝐴) |
|
Theorem | nndomo 6858 |
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 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 ∈ (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 6846 through phplem4 6849, nneneq 6851, 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 6860 |
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 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 𝐵 → 𝐴 ≈ 𝐵)) |
|
2.6.31 Finite sets
|
|
Theorem | fict 6862 |
A finite set is dominated by ω. Also see finct 7109. (Contributed
by Thierry Arnoux, 27-Mar-2018.)
|
⊢ (𝐴 ∈ Fin → 𝐴 ≼ ω) |
|
Theorem | fidceq 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 ∧ 𝐵 ∈ 𝐴 ∧ 𝐶 ∈ 𝐴) → DECID 𝐵 = 𝐶) |
|
Theorem | fidifsnen 6864 |
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 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 ∧ 𝐵 ∈ 𝐴) → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) = 𝐴) |
|
Theorem | nnfi 6866 |
Natural numbers are finite sets. (Contributed by Stefan O'Rear,
21-Mar-2015.)
|
⊢ (𝐴 ∈ ω → 𝐴 ∈ Fin) |
|
Theorem | enfi 6867 |
Equinumerous sets have the same finiteness. (Contributed by NM,
22-Aug-2008.)
|
⊢ (𝐴 ≈ 𝐵 → (𝐴 ∈ Fin ↔ 𝐵 ∈ Fin)) |
|
Theorem | enfii 6868 |
A set equinumerous to a finite set is finite. (Contributed by Mario
Carneiro, 12-Mar-2015.)
|
⊢ ((𝐵 ∈ Fin ∧ 𝐴 ≈ 𝐵) → 𝐴 ∈ Fin) |
|
Theorem | ssfilem 6869* |
Lemma for ssfiexmid 6870. (Contributed by Jim Kingdon, 3-Feb-2022.)
|
⊢ {𝑧 ∈ {∅} ∣ 𝜑} ∈ Fin ⇒ ⊢ (𝜑 ∨ ¬ 𝜑) |
|
Theorem | ssfiexmid 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.)
|
⊢ ∀𝑥∀𝑦((𝑥 ∈ Fin ∧ 𝑦 ⊆ 𝑥) → 𝑦 ∈ Fin) ⇒ ⊢ (𝜑 ∨ ¬ 𝜑) |
|
Theorem | infiexmid 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 → (𝑥 ∩ 𝑦) ∈ Fin) ⇒ ⊢ (𝜑 ∨ ¬ 𝜑) |
|
Theorem | domfiexmid 6872* |
If any set dominated by a finite set is finite, excluded middle follows.
(Contributed by Jim Kingdon, 3-Feb-2022.)
|
⊢ ((𝑥 ∈ Fin ∧ 𝑦 ≼ 𝑥) → 𝑦 ∈ Fin) ⇒ ⊢ (𝜑 ∨ ¬ 𝜑) |
|
Theorem | dif1en 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 𝑀 ∧ 𝑋 ∈ 𝐴) → (𝐴 ∖ {𝑋}) ≈ 𝑀) |
|
Theorem | dif1enen 6874 |
Subtracting one element from each of two equinumerous finite sets.
(Contributed by Jim Kingdon, 5-Jun-2022.)
|
⊢ (𝜑 → 𝐴 ∈ Fin) & ⊢ (𝜑 → 𝐴 ≈ 𝐵)
& ⊢ (𝜑 → 𝐶 ∈ 𝐴)
& ⊢ (𝜑 → 𝐷 ∈ 𝐵) ⇒ ⊢ (𝜑 → (𝐴 ∖ {𝐶}) ≈ (𝐵 ∖ {𝐷})) |
|
Theorem | fiunsnnn 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 ∧ 𝐴 ≈ 𝑁)) → (𝐴 ∪ {𝐵}) ≈ suc 𝑁) |
|
Theorem | php5fin 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})) |
|
Theorem | fisbth 6877 |
Schroeder-Bernstein Theorem for finite sets. (Contributed by Jim
Kingdon, 12-Sep-2021.)
|
⊢ (((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) ∧ (𝐴 ≼ 𝐵 ∧ 𝐵 ≼ 𝐴)) → 𝐴 ≈ 𝐵) |
|
Theorem | 0fin 6878 |
The empty set is finite. (Contributed by FL, 14-Jul-2008.)
|
⊢ ∅ ∈ Fin |
|
Theorem | fin0 6879* |
A nonempty finite set has at least one element. (Contributed by Jim
Kingdon, 10-Sep-2021.)
|
⊢ (𝐴 ∈ Fin → (𝐴 ≠ ∅ ↔ ∃𝑥 𝑥 ∈ 𝐴)) |
|
Theorem | fin0or 6880* |
A finite set is either empty or inhabited. (Contributed by Jim Kingdon,
30-Sep-2021.)
|
⊢ (𝐴 ∈ Fin → (𝐴 = ∅ ∨ ∃𝑥 𝑥 ∈ 𝐴)) |
|
Theorem | diffitest 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 → (𝐴 ∖ 𝐵) ∈ Fin. (Contributed by Jim
Kingdon,
8-Sep-2021.)
|
⊢ ∀𝑎 ∈ Fin ∀𝑏(𝑎 ∖ 𝑏) ∈ Fin ⇒ ⊢ (¬ 𝜑 ∨ ¬ ¬ 𝜑) |
|
Theorem | findcard 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 = ∅ → (𝜑 ↔ 𝜓)) & ⊢ (𝑥 = (𝑦 ∖ {𝑧}) → (𝜑 ↔ 𝜒)) & ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜃)) & ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜏)) & ⊢ 𝜓 & ⊢ (𝑦 ∈ Fin →
(∀𝑧 ∈ 𝑦 𝜒 → 𝜃)) ⇒ ⊢ (𝐴 ∈ Fin → 𝜏) |
|
Theorem | findcard2 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 = ∅ → (𝜑 ↔ 𝜓)) & ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜒)) & ⊢ (𝑥 = (𝑦 ∪ {𝑧}) → (𝜑 ↔ 𝜃)) & ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜏)) & ⊢ 𝜓 & ⊢ (𝑦 ∈ Fin → (𝜒 → 𝜃)) ⇒ ⊢ (𝐴 ∈ Fin → 𝜏) |
|
Theorem | findcard2s 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 = ∅ → (𝜑 ↔ 𝜓)) & ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜒)) & ⊢ (𝑥 = (𝑦 ∪ {𝑧}) → (𝜑 ↔ 𝜃)) & ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜏)) & ⊢ 𝜓 & ⊢ ((𝑦 ∈ Fin ∧ ¬ 𝑧 ∈ 𝑦) → (𝜒 → 𝜃)) ⇒ ⊢ (𝐴 ∈ Fin → 𝜏) |
|
Theorem | findcard2d 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 = ∅ → (𝜓 ↔ 𝜒)) & ⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜃)) & ⊢ (𝑥 = (𝑦 ∪ {𝑧}) → (𝜓 ↔ 𝜏)) & ⊢ (𝑥 = 𝐴 → (𝜓 ↔ 𝜂)) & ⊢ (𝜑 → 𝜒)
& ⊢ ((𝜑 ∧ (𝑦 ⊆ 𝐴 ∧ 𝑧 ∈ (𝐴 ∖ 𝑦))) → (𝜃 → 𝜏)) & ⊢ (𝜑 → 𝐴 ∈ Fin) ⇒ ⊢ (𝜑 → 𝜂) |
|
Theorem | findcard2sd 6886* |
Deduction form of finite set induction . (Contributed by Jim Kingdon,
14-Sep-2021.)
|
⊢ (𝑥 = ∅ → (𝜓 ↔ 𝜒)) & ⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜃)) & ⊢ (𝑥 = (𝑦 ∪ {𝑧}) → (𝜓 ↔ 𝜏)) & ⊢ (𝑥 = 𝐴 → (𝜓 ↔ 𝜂)) & ⊢ (𝜑 → 𝜒)
& ⊢ (((𝜑 ∧ 𝑦 ∈ Fin) ∧ (𝑦 ⊆ 𝐴 ∧ 𝑧 ∈ (𝐴 ∖ 𝑦))) → (𝜃 → 𝜏)) & ⊢ (𝜑 → 𝐴 ∈ Fin) ⇒ ⊢ (𝜑 → 𝜂) |
|
Theorem | diffisn 6887 |
Subtracting a singleton from a finite set produces a finite set.
(Contributed by Jim Kingdon, 11-Sep-2021.)
|
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ 𝐴) → (𝐴 ∖ {𝐵}) ∈ Fin) |
|
Theorem | diffifi 6888 |
Subtracting one finite set from another produces a finite set.
(Contributed by Jim Kingdon, 8-Sep-2021.)
|
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin ∧ 𝐵 ⊆ 𝐴) → (𝐴 ∖ 𝐵) ∈ Fin) |
|
Theorem | infnfi 6889 |
An infinite set is not finite. (Contributed by Jim Kingdon,
20-Feb-2022.)
|
⊢ (ω ≼ 𝐴 → ¬ 𝐴 ∈ Fin) |
|
Theorem | ominf 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 |
|
Theorem | isinfinf 6891* |
An infinite set contains subsets of arbitrarily large finite
cardinality. (Contributed by Jim Kingdon, 15-Jun-2022.)
|
⊢ (ω ≼ 𝐴 → ∀𝑛 ∈ ω ∃𝑥(𝑥 ⊆ 𝐴 ∧ 𝑥 ≈ 𝑛)) |
|
Theorem | ac6sfi 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 = (𝑓‘𝑥) → (𝜑 ↔ 𝜓)) ⇒ ⊢ ((𝐴 ∈ Fin ∧ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝜑) → ∃𝑓(𝑓:𝐴⟶𝐵 ∧ ∀𝑥 ∈ 𝐴 𝜓)) |
|
Theorem | tridc 6893* |
A trichotomous order is decidable. (Contributed by Jim Kingdon,
5-Sep-2022.)
|
⊢ (𝜑 → 𝑅 Po 𝐴)
& ⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥))
& ⊢ (𝜑 → 𝐵 ∈ 𝐴)
& ⊢ (𝜑 → 𝐶 ∈ 𝐴) ⇒ ⊢ (𝜑 → DECID 𝐵𝑅𝐶) |
|
Theorem | fimax2gtrilemstep 6894* |
Lemma for fimax2gtri 6895. The induction step. (Contributed by Jim
Kingdon, 5-Sep-2022.)
|
⊢ (𝜑 → 𝑅 Po 𝐴)
& ⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥))
& ⊢ (𝜑 → 𝐴 ∈ Fin) & ⊢ (𝜑 → 𝐴 ≠ ∅) & ⊢ (𝜑 → 𝑈 ∈ Fin) & ⊢ (𝜑 → 𝑈 ⊆ 𝐴)
& ⊢ (𝜑 → 𝑍 ∈ 𝐴)
& ⊢ (𝜑 → 𝑉 ∈ 𝐴)
& ⊢ (𝜑 → ¬ 𝑉 ∈ 𝑈)
& ⊢ (𝜑 → ∀𝑦 ∈ 𝑈 ¬ 𝑍𝑅𝑦) ⇒ ⊢ (𝜑 → ∃𝑥 ∈ 𝐴 ∀𝑦 ∈ (𝑈 ∪ {𝑉}) ¬ 𝑥𝑅𝑦) |
|
Theorem | fimax2gtri 6895* |
A finite set has a maximum under a trichotomous order. (Contributed
by Jim Kingdon, 5-Sep-2022.)
|
⊢ (𝜑 → 𝑅 Po 𝐴)
& ⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥))
& ⊢ (𝜑 → 𝐴 ∈ Fin) & ⊢ (𝜑 → 𝐴 ≠ ∅)
⇒ ⊢ (𝜑 → ∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥𝑅𝑦) |
|
Theorem | finexdc 6896* |
Decidability of existence, over a finite set and defined by a decidable
proposition. (Contributed by Jim Kingdon, 12-Jul-2022.)
|
⊢ ((𝐴 ∈ Fin ∧ ∀𝑥 ∈ 𝐴 DECID 𝜑) → DECID ∃𝑥 ∈ 𝐴 𝜑) |
|
Theorem | dfrex2fin 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 ∧ ∀𝑥 ∈ 𝐴 DECID 𝜑) → (∃𝑥 ∈ 𝐴 𝜑 ↔ ¬ ∀𝑥 ∈ 𝐴 ¬ 𝜑)) |
|
Theorem | infm 6898* |
An infinite set is inhabited. (Contributed by Jim Kingdon,
18-Feb-2022.)
|
⊢ (ω ≼ 𝐴 → ∃𝑥 𝑥 ∈ 𝐴) |
|
Theorem | infn0 6899 |
An infinite set is not empty. (Contributed by NM, 23-Oct-2004.)
|
⊢ (ω ≼ 𝐴 → 𝐴 ≠ ∅) |
|
Theorem | inffiexmid 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) |