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Theorem List for Intuitionistic Logic Explorer - 6501-6600   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremxpdom1g 6501 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 6502* 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 6503 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 6504 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 6505 Any set dominates the empty set. (Contributed by NM, 26-Oct-2003.) (Revised by Mario Carneiro, 26-Apr-2015.)
(𝐴𝑉 → ∅ ≼ 𝐴)
 
Theoremdom0 6506 A set dominated by the empty set is empty. (Contributed by NM, 22-Nov-2004.)
(𝐴 ≼ ∅ ↔ 𝐴 = ∅)
 
Theorem0dom 6507 Any set dominates the empty set. (Contributed by NM, 26-Oct-2003.) (Revised by Mario Carneiro, 26-Apr-2015.)
𝐴 ∈ V       ∅ ≼ 𝐴
 
Theoremenen1 6508 Equality-like theorem for equinumerosity. (Contributed by NM, 18-Dec-2003.)
(𝐴𝐵 → (𝐴𝐶𝐵𝐶))
 
Theoremenen2 6509 Equality-like theorem for equinumerosity. (Contributed by NM, 18-Dec-2003.)
(𝐴𝐵 → (𝐶𝐴𝐶𝐵))
 
Theoremdomen1 6510 Equality-like theorem for equinumerosity and dominance. (Contributed by NM, 8-Nov-2003.)
(𝐴𝐵 → (𝐴𝐶𝐵𝐶))
 
Theoremdomen2 6511 Equality-like theorem for equinumerosity and dominance. (Contributed by NM, 8-Nov-2003.)
(𝐴𝐵 → (𝐶𝐴𝐶𝐵))
 
2.6.27  Equinumerosity (cont.)
 
Theoremxpf1o 6512* 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 6513 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 6514 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 6515 Order-preserving property of set exponentiation. (Contributed by Jim Kingdon, 15-Jul-2022.)
((𝐴𝐵𝐶𝑉) → (𝐴𝑚 𝐶) ≼ (𝐵𝑚 𝐶))
 
Theoremmapxpen 6516 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 6517* Lemma for xpmapen 6518. (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 6518 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 6519* 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)})
 
2.6.28  Pigeonhole Principle
 
Theoremphplem1 6520 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 6521 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 6522 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 6524. (Contributed by NM, 26-May-1998.)
𝐴 ∈ V    &   𝐵 ∈ V       ((𝐴 ∈ ω ∧ 𝐵 ∈ suc 𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵}))
 
Theoremphplem4 6523 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 6524 A natural number is equinumerous to its successor minus any element of the successor. Version of phplem3 6522 with unnecessary hypotheses removed. (Contributed by Jim Kingdon, 1-Sep-2021.)
((𝐴 ∈ ω ∧ 𝐵 ∈ suc 𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵}))
 
Theoremnneneq 6525 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 6526 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 6527 A singleton {𝐴} is never equinumerous with the ordinal number 2. If 𝐴 is a proper class, see snnen2oprc 6528. (Contributed by Jim Kingdon, 1-Sep-2021.)
(𝐴𝑉 → ¬ {𝐴} ≈ 2𝑜)
 
Theoremsnnen2oprc 6528 A singleton {𝐴} is never equinumerous with the ordinal number 2. If 𝐴 is a set, see snnen2og 6527. (Contributed by Jim Kingdon, 1-Sep-2021.)
𝐴 ∈ V → ¬ {𝐴} ≈ 2𝑜)
 
Theorem1nen2 6529 One and two are not equinumerous. (Contributed by Jim Kingdon, 25-Jan-2022.)
¬ 1𝑜 ≈ 2𝑜
 
Theoremphplem4dom 6530 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 6531 A natural number does not dominate its successor. (Contributed by Jim Kingdon, 1-Sep-2021.)
(𝐴 ∈ ω → ¬ suc 𝐴𝐴)
 
Theoremnndomo 6532 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 6533* 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 6520 through phplem4 6523, nneneq 6525, and this final piece of the proof. (Contributed by NM, 29-May-1998.)
((𝐴 ∈ ω ∧ 𝐵𝐴 ∧ ∃𝑥 𝑥 ∈ (𝐴𝐵)) → ¬ 𝐴𝐵)
 
Theoremphpelm 6534 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 6535 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))
 
2.6.29  Finite sets
 
Theoremfict 6536 A finite set is countable. (Contributed by Thierry Arnoux, 27-Mar-2018.)
(𝐴 ∈ Fin → 𝐴 ≼ ω)
 
Theoremfidceq 6537 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 6538 All decrements of a finite set are equinumerous. (Contributed by Jim Kingdon, 9-Sep-2021.)
((𝑋 ∈ Fin ∧ 𝐴𝑋𝐵𝑋) → (𝑋 ∖ {𝐴}) ≈ (𝑋 ∖ {𝐵}))
 
Theoremfidifsnid 6539 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 3566 from subset to equality when the set is finite. (Contributed by Jim Kingdon, 9-Sep-2021.)
((𝐴 ∈ Fin ∧ 𝐵𝐴) → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) = 𝐴)
 
Theoremnnfi 6540 Natural numbers are finite sets. (Contributed by Stefan O'Rear, 21-Mar-2015.)
(𝐴 ∈ ω → 𝐴 ∈ Fin)
 
Theoremenfi 6541 Equinumerous sets have the same finiteness. (Contributed by NM, 22-Aug-2008.)
(𝐴𝐵 → (𝐴 ∈ Fin ↔ 𝐵 ∈ Fin))
 
Theoremenfii 6542 A set equinumerous to a finite set is finite. (Contributed by Mario Carneiro, 12-Mar-2015.)
((𝐵 ∈ Fin ∧ 𝐴𝐵) → 𝐴 ∈ Fin)
 
Theoremssfilem 6543* Lemma for ssfiexmid 6544. (Contributed by Jim Kingdon, 3-Feb-2022.)
{𝑧 ∈ {∅} ∣ 𝜑} ∈ Fin       (𝜑 ∨ ¬ 𝜑)
 
Theoremssfiexmid 6544* 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 6545* 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 6546* 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 6547 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 6548 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 6549 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 6550 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 6551 Schroeder-Bernstein Theorem for finite sets. (Contributed by Jim Kingdon, 12-Sep-2021.)
(((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) ∧ (𝐴𝐵𝐵𝐴)) → 𝐴𝐵)
 
Theorem0fin 6552 The empty set is finite. (Contributed by FL, 14-Jul-2008.)
∅ ∈ Fin
 
Theoremfin0 6553* A nonempty finite set has at least one element. (Contributed by Jim Kingdon, 10-Sep-2021.)
(𝐴 ∈ Fin → (𝐴 ≠ ∅ ↔ ∃𝑥 𝑥𝐴))
 
Theoremfin0or 6554* A finite set is either empty or inhabited. (Contributed by Jim Kingdon, 30-Sep-2021.)
(𝐴 ∈ Fin → (𝐴 = ∅ ∨ ∃𝑥 𝑥𝐴))
 
Theoremdiffitest 6555* 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 6556* 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 6557* 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 6558* Variation of findcard2 6557 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 6559* Deduction version of findcard2 6557. If you also need 𝑦 ∈ Fin (which doesn't come for free due to ssfiexmid 6544), use findcard2sd 6560 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 6560* 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 6561 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 6562 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 6563 An infinite set is not finite. (Contributed by Jim Kingdon, 20-Feb-2022.)
(ω ≼ 𝐴 → ¬ 𝐴 ∈ Fin)
 
Theoremominf 6564 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 6436. (Contributed by NM, 2-Jun-1998.)
¬ ω ∈ Fin
 
Theoremisinfinf 6565* An infinite set contains subsets of arbitrarily large finite cardinality. (Contributed by Jim Kingdon, 15-Jun-2022.)
(ω ≼ 𝐴 → ∀𝑛 ∈ ω ∃𝑥(𝑥𝐴𝑥𝑛))
 
Theoremac6sfi 6566* 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 6567* A trichotomous order is decidable. (Contributed by Jim Kingdon, 5-Sep-2022.)
(𝜑𝑅 Po 𝐴)    &   (𝜑 → ∀𝑥𝐴𝑦𝐴 (𝑥𝑅𝑦𝑥 = 𝑦𝑦𝑅𝑥))    &   (𝜑𝐵𝐴)    &   (𝜑𝐶𝐴)       (𝜑DECID 𝐵𝑅𝐶)
 
Theoremfimax2gtrilemstep 6568* Lemma for fimax2gtri 6569. 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 6569* 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 6570* 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 6571* 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 6572* An infinite set is inhabited. (Contributed by Jim Kingdon, 18-Feb-2022.)
(ω ≼ 𝐴 → ∃𝑥 𝑥𝐴)
 
Theoreminfn0 6573 An infinite set is not empty. (Contributed by NM, 23-Oct-2004.)
(ω ≼ 𝐴𝐴 ≠ ∅)
 
Theoreminffiexmid 6574* 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)
 
Theoremen2eqpr 6575 Building a set with two elements. (Contributed by FL, 11-Aug-2008.) (Revised by Mario Carneiro, 10-Sep-2015.)
((𝐶 ≈ 2𝑜𝐴𝐶𝐵𝐶) → (𝐴𝐵𝐶 = {𝐴, 𝐵}))
 
Theoremexmidpw 6576 Excluded middle is equivalent to the power set of 1𝑜 having two elements. Remark of [PradicBrown2022], p. 2. (Contributed by Jim Kingdon, 30-Jun-2022.)
(EXMID ↔ 𝒫 1𝑜 ≈ 2𝑜)
 
Theoremfientri3 6577 Trichotomy of dominance for finite sets. (Contributed by Jim Kingdon, 15-Sep-2021.)
((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) → (𝐴𝐵𝐵𝐴))
 
Theoremnnwetri 6578* A natural number is well-ordered by E. More specifically, this order both satisfies We and is trichotomous. (Contributed by Jim Kingdon, 25-Sep-2021.)
(𝐴 ∈ ω → ( E We 𝐴 ∧ ∀𝑥𝐴𝑦𝐴 (𝑥 E 𝑦𝑥 = 𝑦𝑦 E 𝑥)))
 
Theoremonunsnss 6579 Adding a singleton to create an ordinal. (Contributed by Jim Kingdon, 20-Oct-2021.)
((𝐵𝑉 ∧ (𝐴 ∪ {𝐵}) ∈ On) → 𝐵𝐴)
 
Theoremunfiexmid 6580* If the union of any two finite sets is finite, excluded middle follows. Remark 8.1.17 of [AczelRathjen], p. 74. (Contributed by Mario Carneiro and Jim Kingdon, 5-Mar-2022.)
((𝑥 ∈ Fin ∧ 𝑦 ∈ Fin) → (𝑥𝑦) ∈ Fin)       (𝜑 ∨ ¬ 𝜑)
 
Theoremunsnfi 6581 Adding a singleton to a finite set yields a finite set. (Contributed by Jim Kingdon, 3-Feb-2022.)
((𝐴 ∈ Fin ∧ 𝐵𝑉 ∧ ¬ 𝐵𝐴) → (𝐴 ∪ {𝐵}) ∈ Fin)
 
Theoremunsnfidcex 6582 The 𝐵𝑉 condition in unsnfi 6581. This is intended to show that unsnfi 6581 without that condition would not be provable but it probably would need to be strengthened (for example, to imply included middle) to fully show that. (Contributed by Jim Kingdon, 6-Feb-2022.)
((𝐴 ∈ Fin ∧ ¬ 𝐵𝐴 ∧ (𝐴 ∪ {𝐵}) ∈ Fin) → DECID ¬ 𝐵 ∈ V)
 
Theoremunsnfidcel 6583 The ¬ 𝐵𝐴 condition in unsnfi 6581. This is intended to show that unsnfi 6581 without that condition would not be provable but it probably would need to be strengthened (for example, to imply included middle) to fully show that. (Contributed by Jim Kingdon, 6-Feb-2022.)
((𝐴 ∈ Fin ∧ 𝐵𝑉 ∧ (𝐴 ∪ {𝐵}) ∈ Fin) → DECID ¬ 𝐵𝐴)
 
Theoremunfidisj 6584 The union of two disjoint finite sets is finite. (Contributed by Jim Kingdon, 25-Feb-2022.)
((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin ∧ (𝐴𝐵) = ∅) → (𝐴𝐵) ∈ Fin)
 
Theoremundifdcss 6585* Union of complementary parts into whole and decidability. (Contributed by Jim Kingdon, 17-Jun-2022.)
(𝐴 = (𝐵 ∪ (𝐴𝐵)) ↔ (𝐵𝐴 ∧ ∀𝑥𝐴 DECID 𝑥𝐵))
 
Theoremundifdc 6586* Union of complementary parts into whole. This is a case where we can strengthen undifss 3350 from subset to equality. (Contributed by Jim Kingdon, 17-Jun-2022.)
((∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦𝐵 ∈ Fin ∧ 𝐵𝐴) → 𝐴 = (𝐵 ∪ (𝐴𝐵)))
 
Theoremundiffi 6587 Union of complementary parts into whole. This is a case where we can strengthen undifss 3350 from subset to equality. (Contributed by Jim Kingdon, 2-Mar-2022.)
((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin ∧ 𝐵𝐴) → 𝐴 = (𝐵 ∪ (𝐴𝐵)))
 
Theoremunfiin 6588 The union of two finite sets is finite if their intersection is. (Contributed by Jim Kingdon, 2-Mar-2022.)
((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin ∧ (𝐴𝐵) ∈ Fin) → (𝐴𝐵) ∈ Fin)
 
Theoremprfidisj 6589 A pair is finite if it consists of two unequal sets. For the case where 𝐴 = 𝐵, see snfig 6483. For the cases where one or both is a proper class, see prprc1 3533, prprc2 3534, or prprc 3535. (Contributed by Jim Kingdon, 31-May-2022.)
((𝐴𝑉𝐵𝑊𝐴𝐵) → {𝐴, 𝐵} ∈ Fin)
 
Theoremxpfi 6590 The Cartesian product of two finite sets is finite. Lemma 8.1.16 of [AczelRathjen], p. 74. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 12-Mar-2015.)
((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) → (𝐴 × 𝐵) ∈ Fin)
 
Theorem3xpfi 6591 The Cartesian product of three finite sets is a finite set. (Contributed by Alexander van der Vekens, 11-Mar-2018.)
(𝑉 ∈ Fin → ((𝑉 × 𝑉) × 𝑉) ∈ Fin)
 
Theoremfisseneq 6592 A finite set is equal to its subset if they are equinumerous. (Contributed by FL, 11-Aug-2008.)
((𝐵 ∈ Fin ∧ 𝐴𝐵𝐴𝐵) → 𝐴 = 𝐵)
 
Theoremssfirab 6593* A subset of a finite set is finite if it is defined by a decidable property. (Contributed by Jim Kingdon, 27-May-2022.)
(𝜑𝐴 ∈ Fin)    &   (𝜑 → ∀𝑥𝐴 DECID 𝜓)       (𝜑 → {𝑥𝐴𝜓} ∈ Fin)
 
Theoremssfidc 6594* A subset of a finite set is finite if membership in the subset is decidable. (Contributed by Jim Kingdon, 27-May-2022.)
((𝐴 ∈ Fin ∧ 𝐵𝐴 ∧ ∀𝑥𝐴 DECID 𝑥𝐵) → 𝐵 ∈ Fin)
 
Theoremsnon0 6595 An ordinal which is a singleton is {∅}. (Contributed by Jim Kingdon, 19-Oct-2021.)
((𝐴𝑉 ∧ {𝐴} ∈ On) → 𝐴 = ∅)
 
Theoremfnfi 6596 A version of fnex 5480 for finite sets. (Contributed by Mario Carneiro, 16-Nov-2014.) (Revised by Mario Carneiro, 24-Jun-2015.)
((𝐹 Fn 𝐴𝐴 ∈ Fin) → 𝐹 ∈ Fin)
 
Theoremfundmfi 6597 The domain of a finite function is finite. (Contributed by Jim Kingdon, 5-Feb-2022.)
((𝐴 ∈ Fin ∧ Fun 𝐴) → dom 𝐴 ∈ Fin)
 
Theoremfundmfibi 6598 A function is finite if and only if its domain is finite. (Contributed by AV, 10-Jan-2020.)
(Fun 𝐹 → (𝐹 ∈ Fin ↔ dom 𝐹 ∈ Fin))
 
Theoremresfnfinfinss 6599 The restriction of a function to a finite subset of its domain is finite. (Contributed by Alexander van der Vekens, 3-Feb-2018.)
((𝐹 Fn 𝐴𝐵 ∈ Fin ∧ 𝐵𝐴) → (𝐹𝐵) ∈ Fin)
 
Theoremrelcnvfi 6600 If a relation is finite, its converse is as well. (Contributed by Jim Kingdon, 5-Feb-2022.)
((Rel 𝐴𝐴 ∈ Fin) → 𝐴 ∈ Fin)
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