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Theorem List for Intuitionistic Logic Explorer - 6301-6400   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremdomentr 6301 Transitivity of dominance and equinumerosity. (Contributed by NM, 7-Jun-1998.)
((𝐴𝐵𝐵𝐶) → 𝐴𝐶)
 
Theoremf1imaeng 6302 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:𝐴1-1𝐵𝐶𝐴𝐶𝑉) → (𝐹𝐶) ≈ 𝐶)
 
Theoremf1imaen2g 6303 A one-to-one function's image under a subset of its domain is equinumerous to the subset. (This version of f1imaen 6304 does not need ax-setind 4289.) (Contributed by Mario Carneiro, 16-Nov-2014.) (Revised by Mario Carneiro, 25-Jun-2015.)
(((𝐹:𝐴1-1𝐵𝐵𝑉) ∧ (𝐶𝐴𝐶𝑉)) → (𝐹𝐶) ≈ 𝐶)
 
Theoremf1imaen 6304 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:𝐴1-1𝐵𝐶𝐴) → (𝐹𝐶) ≈ 𝐶)
 
Theoremen0 6305 The empty set is equinumerous only to itself. Exercise 1 of [TakeutiZaring] p. 88. (Contributed by NM, 27-May-1998.)
(𝐴 ≈ ∅ ↔ 𝐴 = ∅)
 
Theoremensn1 6306 A singleton is equinumerous to ordinal one. (Contributed by NM, 4-Nov-2002.)
𝐴 ∈ V       {𝐴} ≈ 1𝑜
 
Theoremensn1g 6307 A singleton is equinumerous to ordinal one. (Contributed by NM, 23-Apr-2004.)
(𝐴𝑉 → {𝐴} ≈ 1𝑜)
 
Theoremenpr1g 6308 {𝐴, 𝐴} has only one element. (Contributed by FL, 15-Feb-2010.)
(𝐴𝑉 → {𝐴, 𝐴} ≈ 1𝑜)
 
Theoremen1 6309* A set is equinumerous to ordinal one iff it is a singleton. (Contributed by NM, 25-Jul-2004.)
(𝐴 ≈ 1𝑜 ↔ ∃𝑥 𝐴 = {𝑥})
 
Theoremen1bg 6310 A set is equinumerous to ordinal one iff it is a singleton. (Contributed by Jim Kingdon, 13-Apr-2020.)
(𝐴𝑉 → (𝐴 ≈ 1𝑜𝐴 = { 𝐴}))
 
Theoremreuen1 6311* Two ways to express "exactly one". (Contributed by Stefan O'Rear, 28-Oct-2014.)
(∃!𝑥𝐴 𝜑 ↔ {𝑥𝐴𝜑} ≈ 1𝑜)
 
Theoremeuen1 6312 Two ways to express "exactly one". (Contributed by Stefan O'Rear, 28-Oct-2014.)
(∃!𝑥𝜑 ↔ {𝑥𝜑} ≈ 1𝑜)
 
Theoremeuen1b 6313* Two ways to express "𝐴 has a unique element". (Contributed by Mario Carneiro, 9-Apr-2015.)
(𝐴 ≈ 1𝑜 ↔ ∃!𝑥 𝑥𝐴)
 
Theoremen1uniel 6314 A singleton contains its sole element. (Contributed by Stefan O'Rear, 16-Aug-2015.)
(𝑆 ≈ 1𝑜 𝑆𝑆)
 
Theorem2dom 6315* A set that dominates ordinal 2 has at least 2 different members. (Contributed by NM, 25-Jul-2004.)
(2𝑜𝐴 → ∃𝑥𝐴𝑦𝐴 ¬ 𝑥 = 𝑦)
 
Theoremfundmen 6316 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 6317 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 6318 A relational set is equinumerous to its converse. (Contributed by Mario Carneiro, 28-Dec-2014.)
((Rel 𝐴𝐴𝑉) → 𝐴𝐴)
 
Theoremfndmeng 6319 A function is equinumerate to its domain. (Contributed by Paul Chapman, 22-Jun-2011.)
((𝐹 Fn 𝐴𝐴𝐶) → 𝐴𝐹)
 
Theoremen2sn 6320 Two singletons are equinumerous. (Contributed by NM, 9-Nov-2003.)
((𝐴𝐶𝐵𝐷) → {𝐴} ≈ {𝐵})
 
Theoremsnfig 6321 A singleton is finite. (Contributed by Jim Kingdon, 13-Apr-2020.)
(𝐴𝑉 → {𝐴} ∈ Fin)
 
Theoremfiprc 6322 The class of finite sets is a proper class. (Contributed by Jeff Hankins, 3-Oct-2008.)
Fin ∉ V
 
Theoremunen 6323 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))
 
Theoremenm 6324* A set equinumerous to an inhabited set is inhabited. (Contributed by Jim Kingdon, 19-May-2020.)
((𝐴𝐵 ∧ ∃𝑥 𝑥𝐴) → ∃𝑦 𝑦𝐵)
 
Theoremxpsnen 6325 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 6326 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 6327 One times a cardinal number. (Contributed by NM, 27-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
(𝐴𝑉 → (𝐴 × 1𝑜) ≈ 𝐴)
 
Theoremendisj 6328* 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 6329* 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 6330* Composition with the bijection of xpcomf1o 6329 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 6331 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 6332 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 6333 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 6334 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 6335 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 6336 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 6337 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 6338* 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 6339 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 6340 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)
 
Theoremenen1 6341 Equality-like theorem for equinumerosity. (Contributed by NM, 18-Dec-2003.)
(𝐴𝐵 → (𝐴𝐶𝐵𝐶))
 
Theoremenen2 6342 Equality-like theorem for equinumerosity. (Contributed by NM, 18-Dec-2003.)
(𝐴𝐵 → (𝐶𝐴𝐶𝐵))
 
Theoremdomen1 6343 Equality-like theorem for equinumerosity and dominance. (Contributed by NM, 8-Nov-2003.)
(𝐴𝐵 → (𝐴𝐶𝐵𝐶))
 
Theoremdomen2 6344 Equality-like theorem for equinumerosity and dominance. (Contributed by NM, 8-Nov-2003.)
(𝐴𝐵 → (𝐶𝐴𝐶𝐵))
 
2.6.26  Pigeonhole Principle
 
Theoremphplem1 6345 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 6346 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 6347 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 6349. (Contributed by NM, 26-May-1998.)
𝐴 ∈ V    &   𝐵 ∈ V       ((𝐴 ∈ ω ∧ 𝐵 ∈ suc 𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵}))
 
Theoremphplem4 6348 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 6349 A natural number is equinumerous to its successor minus any element of the successor. Version of phplem3 6347 with unnecessary hypotheses removed. (Contributed by Jim Kingdon, 1-Sep-2021.)
((𝐴 ∈ ω ∧ 𝐵 ∈ suc 𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵}))
 
Theoremnneneq 6350 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 6351 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 6352 A singleton {𝐴} is never equinumerous with the ordinal number 2. If 𝐴 is a proper class, see snnen2oprc 6353. (Contributed by Jim Kingdon, 1-Sep-2021.)
(𝐴𝑉 → ¬ {𝐴} ≈ 2𝑜)
 
Theoremsnnen2oprc 6353 A singleton {𝐴} is never equinumerous with the ordinal number 2. If 𝐴 is a set, see snnen2og 6352. (Contributed by Jim Kingdon, 1-Sep-2021.)
𝐴 ∈ V → ¬ {𝐴} ≈ 2𝑜)
 
Theoremphplem4dom 6354 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 6355 A natural number does not dominate its successor. (Contributed by Jim Kingdon, 1-Sep-2021.)
(𝐴 ∈ ω → ¬ suc 𝐴𝐴)
 
Theoremnndomo 6356 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 6357* 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 6345 through phplem4 6348, nneneq 6350, and this final piece of the proof. (Contributed by NM, 29-May-1998.)
((𝐴 ∈ ω ∧ 𝐵𝐴 ∧ ∃𝑥 𝑥 ∈ (𝐴𝐵)) → ¬ 𝐴𝐵)
 
Theoremphpelm 6358 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 6359 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))
 
2.6.27  Finite sets
 
Theoremfidceq 6360 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 6361 All decrements of a finite set are equinumerous. (Contributed by Jim Kingdon, 9-Sep-2021.)
((𝑋 ∈ Fin ∧ 𝐴𝑋𝐵𝑋) → (𝑋 ∖ {𝐴}) ≈ (𝑋 ∖ {𝐵}))
 
Theoremfidifsnid 6362 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 3537 from subset to equality when the set is finite. (Contributed by Jim Kingdon, 9-Sep-2021.)
((𝐴 ∈ Fin ∧ 𝐵𝐴) → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) = 𝐴)
 
Theoremnnfi 6363 Natural numbers are finite sets. (Contributed by Stefan O'Rear, 21-Mar-2015.)
(𝐴 ∈ ω → 𝐴 ∈ Fin)
 
Theoremenfi 6364 Equinumerous sets have the same finiteness. (Contributed by NM, 22-Aug-2008.)
(𝐴𝐵 → (𝐴 ∈ Fin ↔ 𝐵 ∈ Fin))
 
Theoremenfii 6365 A set equinumerous to a finite set is finite. (Contributed by Mario Carneiro, 12-Mar-2015.)
((𝐵 ∈ Fin ∧ 𝐴𝐵) → 𝐴 ∈ Fin)
 
Theoremssfiexmid 6366* 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)
 
Theoremdif1en 6367 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)
 
Theoremfiunsnnn 6368 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 6369 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 6370 Schroeder-Bernstein Theorem for finite sets. (Contributed by Jim Kingdon, 12-Sep-2021.)
(((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) ∧ (𝐴𝐵𝐵𝐴)) → 𝐴𝐵)
 
Theorem0fin 6371 The empty set is finite. (Contributed by FL, 14-Jul-2008.)
∅ ∈ Fin
 
Theoremfin0 6372* A nonempty finite set has at least one element. (Contributed by Jim Kingdon, 10-Sep-2021.)
(𝐴 ∈ Fin → (𝐴 ≠ ∅ ↔ ∃𝑥 𝑥𝐴))
 
Theoremfin0or 6373* A finite set is either empty or inhabited. (Contributed by Jim Kingdon, 30-Sep-2021.)
(𝐴 ∈ Fin → (𝐴 = ∅ ∨ ∃𝑥 𝑥𝐴))
 
Theoremdiffitest 6374* 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 6375* 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 6376* 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 6377* Variation of findcard2 6376 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 6378* Deduction version of findcard2 6376. If you also need 𝑦 ∈ Fin (which doesn't come for free due to ssfiexmid 6366), use findcard2sd 6379 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 6379* 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 6380 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 6381 Subtracting one finite set from another produces a finite set. (Contributed by Jim Kingdon, 8-Sep-2021.)
((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin ∧ 𝐵𝐴) → (𝐴𝐵) ∈ Fin)
 
Theoremac6sfi 6382* 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))
 
Theoremfientri3 6383 Trichotomy of dominance for finite sets. (Contributed by Jim Kingdon, 15-Sep-2021.)
((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) → (𝐴𝐵𝐵𝐴))
 
Theoremnnwetri 6384* 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 6385 Adding a singleton to create an ordinal. (Contributed by Jim Kingdon, 20-Oct-2021.)
((𝐵𝑉 ∧ (𝐴 ∪ {𝐵}) ∈ On) → 𝐵𝐴)
 
Theoremsnon0 6386 An ordinal which is a singleton is {∅}. (Contributed by Jim Kingdon, 19-Oct-2021.)
((𝐴𝑉 ∧ {𝐴} ∈ On) → 𝐴 = ∅)
 
2.6.28  Supremum and infimum
 
Syntaxcsup 6387 Extend class notation to include supremum of class 𝐴. Here 𝑅 is ordinarily a relation that strictly orders class 𝐵. For example, 𝑅 could be 'less than' and 𝐵 could be the set of real numbers.
class sup(𝐴, 𝐵, 𝑅)
 
Syntaxcinf 6388 Extend class notation to include infimum of class 𝐴. Here 𝑅 is ordinarily a relation that strictly orders class 𝐵. For example, 𝑅 could be 'less than' and 𝐵 could be the set of real numbers.
class inf(𝐴, 𝐵, 𝑅)
 
Definitiondf-sup 6389* Define the supremum of class 𝐴. It is meaningful when 𝑅 is a relation that strictly orders 𝐵 and when the supremum exists. (Contributed by NM, 22-May-1999.)
sup(𝐴, 𝐵, 𝑅) = {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐵 (𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧))}
 
Definitiondf-inf 6390 Define the infimum of class 𝐴. It is meaningful when 𝑅 is a relation that strictly orders 𝐵 and when the infimum exists. For example, 𝑅 could be 'less than', 𝐵 could be the set of real numbers, and 𝐴 could be the set of all positive reals; in this case the infimum is 0. The infimum is defined as the supremum using the converse ordering relation. In the given example, 0 is the supremum of all reals (greatest real number) for which all positive reals are greater. (Contributed by AV, 2-Sep-2020.)
inf(𝐴, 𝐵, 𝑅) = sup(𝐴, 𝐵, 𝑅)
 
Theoremsupeq1 6391 Equality theorem for supremum. (Contributed by NM, 22-May-1999.)
(𝐵 = 𝐶 → sup(𝐵, 𝐴, 𝑅) = sup(𝐶, 𝐴, 𝑅))
 
Theoremsupeq1d 6392 Equality deduction for supremum. (Contributed by Paul Chapman, 22-Jun-2011.)
(𝜑𝐵 = 𝐶)       (𝜑 → sup(𝐵, 𝐴, 𝑅) = sup(𝐶, 𝐴, 𝑅))
 
Theoremsupeq1i 6393 Equality inference for supremum. (Contributed by Paul Chapman, 22-Jun-2011.)
𝐵 = 𝐶       sup(𝐵, 𝐴, 𝑅) = sup(𝐶, 𝐴, 𝑅)
 
Theoremsupeq2 6394 Equality theorem for supremum. (Contributed by Jeff Madsen, 2-Sep-2009.)
(𝐵 = 𝐶 → sup(𝐴, 𝐵, 𝑅) = sup(𝐴, 𝐶, 𝑅))
 
Theoremsupeq3 6395 Equality theorem for supremum. (Contributed by Scott Fenton, 13-Jun-2018.)
(𝑅 = 𝑆 → sup(𝐴, 𝐵, 𝑅) = sup(𝐴, 𝐵, 𝑆))
 
Theoremsupeq123d 6396 Equality deduction for supremum. (Contributed by Stefan O'Rear, 20-Jan-2015.)
(𝜑𝐴 = 𝐷)    &   (𝜑𝐵 = 𝐸)    &   (𝜑𝐶 = 𝐹)       (𝜑 → sup(𝐴, 𝐵, 𝐶) = sup(𝐷, 𝐸, 𝐹))
 
Theoremnfsup 6397 Hypothesis builder for supremum. (Contributed by Mario Carneiro, 20-Mar-2014.)
𝑥𝐴    &   𝑥𝐵    &   𝑥𝑅       𝑥sup(𝐴, 𝐵, 𝑅)
 
Theoremsupmoti 6398* Any class 𝐵 has at most one supremum in 𝐴 (where 𝑅 is interpreted as 'less than'). The hypothesis is satisfied by real numbers (see lttri3 7156) or other orders which correspond to tight apartnesses. (Contributed by Jim Kingdon, 23-Nov-2021.)
((𝜑 ∧ (𝑢𝐴𝑣𝐴)) → (𝑢 = 𝑣 ↔ (¬ 𝑢𝑅𝑣 ∧ ¬ 𝑣𝑅𝑢)))       (𝜑 → ∃*𝑥𝐴 (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧)))
 
Theoremsupeuti 6399* A supremum is unique. Similar to Theorem I.26 of [Apostol] p. 24 (but for suprema in general). (Contributed by Jim Kingdon, 23-Nov-2021.)
((𝜑 ∧ (𝑢𝐴𝑣𝐴)) → (𝑢 = 𝑣 ↔ (¬ 𝑢𝑅𝑣 ∧ ¬ 𝑣𝑅𝑢)))    &   (𝜑 → ∃𝑥𝐴 (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧)))       (𝜑 → ∃!𝑥𝐴 (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧)))
 
Theoremsupval2ti 6400* Alternate expression for the supremum. (Contributed by Jim Kingdon, 23-Nov-2021.)
((𝜑 ∧ (𝑢𝐴𝑣𝐴)) → (𝑢 = 𝑣 ↔ (¬ 𝑢𝑅𝑣 ∧ ¬ 𝑣𝑅𝑢)))    &   (𝜑 → ∃𝑥𝐴 (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧)))       (𝜑 → sup(𝐵, 𝐴, 𝑅) = (𝑥𝐴 (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧))))
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