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Theorem List for Intuitionistic Logic Explorer - 5001-5100   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremimaex 5001 The image of a set is a set. Theorem 3.17 of [Monk1] p. 39. (Contributed by JJ, 24-Sep-2021.)
𝐴 ∈ V       (𝐴𝐵) ∈ V
 
Theoremimai 5002 Image under the identity relation. Theorem 3.16(viii) of [Monk1] p. 38. (Contributed by NM, 30-Apr-1998.)
( I “ 𝐴) = 𝐴
 
Theoremrnresi 5003 The range of the restricted identity function. (Contributed by NM, 27-Aug-2004.)
ran ( I ↾ 𝐴) = 𝐴
 
Theoremresiima 5004 The image of a restriction of the identity function. (Contributed by FL, 31-Dec-2006.)
(𝐵𝐴 → (( I ↾ 𝐴) “ 𝐵) = 𝐵)
 
Theoremima0 5005 Image of the empty set. Theorem 3.16(ii) of [Monk1] p. 38. (Contributed by NM, 20-May-1998.)
(𝐴 “ ∅) = ∅
 
Theorem0ima 5006 Image under the empty relation. (Contributed by FL, 11-Jan-2007.)
(∅ “ 𝐴) = ∅
 
Theoremcsbima12g 5007 Move class substitution in and out of the image of a function. (Contributed by FL, 15-Dec-2006.) (Proof shortened by Mario Carneiro, 4-Dec-2016.)
(𝐴𝐶𝐴 / 𝑥(𝐹𝐵) = (𝐴 / 𝑥𝐹𝐴 / 𝑥𝐵))
 
Theoremimadisj 5008 A class whose image under another is empty is disjoint with the other's domain. (Contributed by FL, 24-Jan-2007.)
((𝐴𝐵) = ∅ ↔ (dom 𝐴𝐵) = ∅)
 
Theoremcnvimass 5009 A preimage under any class is included in the domain of the class. (Contributed by FL, 29-Jan-2007.)
(𝐴𝐵) ⊆ dom 𝐴
 
Theoremcnvimarndm 5010 The preimage of the range of a class is the domain of the class. (Contributed by Jeff Hankins, 15-Jul-2009.)
(𝐴 “ ran 𝐴) = dom 𝐴
 
Theoremimasng 5011* The image of a singleton. (Contributed by NM, 8-May-2005.)
(𝐴𝐵 → (𝑅 “ {𝐴}) = {𝑦𝐴𝑅𝑦})
 
Theoremelrelimasn 5012 Elementhood in the image of a singleton. (Contributed by Mario Carneiro, 3-Nov-2015.)
(Rel 𝑅 → (𝐵 ∈ (𝑅 “ {𝐴}) ↔ 𝐴𝑅𝐵))
 
Theoremelimasn 5013 Membership in an image of a singleton. (Contributed by NM, 15-Mar-2004.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
𝐵 ∈ V    &   𝐶 ∈ V       (𝐶 ∈ (𝐴 “ {𝐵}) ↔ ⟨𝐵, 𝐶⟩ ∈ 𝐴)
 
Theoremelimasng 5014 Membership in an image of a singleton. (Contributed by Raph Levien, 21-Oct-2006.)
((𝐵𝑉𝐶𝑊) → (𝐶 ∈ (𝐴 “ {𝐵}) ↔ ⟨𝐵, 𝐶⟩ ∈ 𝐴))
 
Theoremargs 5015* Two ways to express the class of unique-valued arguments of 𝐹, which is the same as the domain of 𝐹 whenever 𝐹 is a function. The left-hand side of the equality is from Definition 10.2 of [Quine] p. 65. Quine uses the notation "arg 𝐹 " for this class (for which we have no separate notation). (Contributed by NM, 8-May-2005.)
{𝑥 ∣ ∃𝑦(𝐹 “ {𝑥}) = {𝑦}} = {𝑥 ∣ ∃!𝑦 𝑥𝐹𝑦}
 
Theoremeliniseg 5016 Membership in an initial segment. The idiom (𝐴 “ {𝐵}), meaning {𝑥𝑥𝐴𝐵}, is used to specify an initial segment in (for example) Definition 6.21 of [TakeutiZaring] p. 30. (Contributed by NM, 28-Apr-2004.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
𝐶 ∈ V       (𝐵𝑉 → (𝐶 ∈ (𝐴 “ {𝐵}) ↔ 𝐶𝐴𝐵))
 
Theoremepini 5017 Any set is equal to its preimage under the converse epsilon relation. (Contributed by Mario Carneiro, 9-Mar-2013.)
𝐴 ∈ V       ( E “ {𝐴}) = 𝐴
 
Theoreminiseg 5018* An idiom that signifies an initial segment of an ordering, used, for example, in Definition 6.21 of [TakeutiZaring] p. 30. (Contributed by NM, 28-Apr-2004.)
(𝐵𝑉 → (𝐴 “ {𝐵}) = {𝑥𝑥𝐴𝐵})
 
Theoremdfse2 5019* Alternate definition of set-like relation. (Contributed by Mario Carneiro, 23-Jun-2015.)
(𝑅 Se 𝐴 ↔ ∀𝑥𝐴 (𝐴 ∩ (𝑅 “ {𝑥})) ∈ V)
 
Theoremexse2 5020 Any set relation is set-like. (Contributed by Mario Carneiro, 22-Jun-2015.)
(𝑅𝑉𝑅 Se 𝐴)
 
Theoremimass1 5021 Subset theorem for image. (Contributed by NM, 16-Mar-2004.)
(𝐴𝐵 → (𝐴𝐶) ⊆ (𝐵𝐶))
 
Theoremimass2 5022 Subset theorem for image. Exercise 22(a) of [Enderton] p. 53. (Contributed by NM, 22-Mar-1998.)
(𝐴𝐵 → (𝐶𝐴) ⊆ (𝐶𝐵))
 
Theoremndmima 5023 The image of a singleton outside the domain is empty. (Contributed by NM, 22-May-1998.)
𝐴 ∈ dom 𝐵 → (𝐵 “ {𝐴}) = ∅)
 
Theoremrelcnv 5024 A converse is a relation. Theorem 12 of [Suppes] p. 62. (Contributed by NM, 29-Oct-1996.)
Rel 𝐴
 
Theoremrelbrcnvg 5025 When 𝑅 is a relation, the sethood assumptions on brcnv 4828 can be omitted. (Contributed by Mario Carneiro, 28-Apr-2015.)
(Rel 𝑅 → (𝐴𝑅𝐵𝐵𝑅𝐴))
 
Theoremeliniseg2 5026 Eliminate the class existence constraint in eliniseg 5016. (Contributed by Mario Carneiro, 5-Dec-2014.) (Revised by Mario Carneiro, 17-Nov-2015.)
(Rel 𝐴 → (𝐶 ∈ (𝐴 “ {𝐵}) ↔ 𝐶𝐴𝐵))
 
Theoremrelbrcnv 5027 When 𝑅 is a relation, the sethood assumptions on brcnv 4828 can be omitted. (Contributed by Mario Carneiro, 28-Apr-2015.)
Rel 𝑅       (𝐴𝑅𝐵𝐵𝑅𝐴)
 
Theoremcotr 5028* Two ways of saying a relation is transitive. Definition of transitivity in [Schechter] p. 51. (Contributed by NM, 27-Dec-1996.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
((𝑅𝑅) ⊆ 𝑅 ↔ ∀𝑥𝑦𝑧((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧))
 
Theoremissref 5029* Two ways to state a relation is reflexive. Adapted from Tarski. (Contributed by FL, 15-Jan-2012.) (Revised by NM, 30-Mar-2016.)
(( I ↾ 𝐴) ⊆ 𝑅 ↔ ∀𝑥𝐴 𝑥𝑅𝑥)
 
Theoremcnvsym 5030* Two ways of saying a relation is symmetric. Similar to definition of symmetry in [Schechter] p. 51. (Contributed by NM, 28-Dec-1996.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
(𝑅𝑅 ↔ ∀𝑥𝑦(𝑥𝑅𝑦𝑦𝑅𝑥))
 
Theoremintasym 5031* Two ways of saying a relation is antisymmetric. Definition of antisymmetry in [Schechter] p. 51. (Contributed by NM, 9-Sep-2004.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
((𝑅𝑅) ⊆ I ↔ ∀𝑥𝑦((𝑥𝑅𝑦𝑦𝑅𝑥) → 𝑥 = 𝑦))
 
Theoremasymref 5032* Two ways of saying a relation is antisymmetric and reflexive. 𝑅 is the field of a relation by relfld 5175. (Contributed by NM, 6-May-2008.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
((𝑅𝑅) = ( I ↾ 𝑅) ↔ ∀𝑥 𝑅𝑦((𝑥𝑅𝑦𝑦𝑅𝑥) ↔ 𝑥 = 𝑦))
 
Theoremintirr 5033* Two ways of saying a relation is irreflexive. Definition of irreflexivity in [Schechter] p. 51. (Contributed by NM, 9-Sep-2004.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
((𝑅 ∩ I ) = ∅ ↔ ∀𝑥 ¬ 𝑥𝑅𝑥)
 
Theorembrcodir 5034* Two ways of saying that two elements have an upper bound. (Contributed by Mario Carneiro, 3-Nov-2015.)
((𝐴𝑉𝐵𝑊) → (𝐴(𝑅𝑅)𝐵 ↔ ∃𝑧(𝐴𝑅𝑧𝐵𝑅𝑧)))
 
Theoremcodir 5035* Two ways of saying a relation is directed. (Contributed by Mario Carneiro, 22-Nov-2013.)
((𝐴 × 𝐵) ⊆ (𝑅𝑅) ↔ ∀𝑥𝐴𝑦𝐵𝑧(𝑥𝑅𝑧𝑦𝑅𝑧))
 
Theoremqfto 5036* A quantifier-free way of expressing the total order predicate. (Contributed by Mario Carneiro, 22-Nov-2013.)
((𝐴 × 𝐵) ⊆ (𝑅𝑅) ↔ ∀𝑥𝐴𝑦𝐵 (𝑥𝑅𝑦𝑦𝑅𝑥))
 
Theoremxpidtr 5037 A square cross product (𝐴 × 𝐴) is a transitive relation. (Contributed by FL, 31-Jul-2009.)
((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) ⊆ (𝐴 × 𝐴)
 
Theoremtrin2 5038 The intersection of two transitive classes is transitive. (Contributed by FL, 31-Jul-2009.)
(((𝑅𝑅) ⊆ 𝑅 ∧ (𝑆𝑆) ⊆ 𝑆) → ((𝑅𝑆) ∘ (𝑅𝑆)) ⊆ (𝑅𝑆))
 
Theorempoirr2 5039 A partial order relation is irreflexive. (Contributed by Mario Carneiro, 2-Nov-2015.)
(𝑅 Po 𝐴 → (𝑅 ∩ ( I ↾ 𝐴)) = ∅)
 
Theoremtrinxp 5040 The relation induced by a transitive relation on a part of its field is transitive. (Taking the intersection of a relation with a square cross product is a way to restrict it to a subset of its field.) (Contributed by FL, 31-Jul-2009.)
((𝑅𝑅) ⊆ 𝑅 → ((𝑅 ∩ (𝐴 × 𝐴)) ∘ (𝑅 ∩ (𝐴 × 𝐴))) ⊆ (𝑅 ∩ (𝐴 × 𝐴)))
 
Theoremsoirri 5041 A strict order relation is irreflexive. (Contributed by NM, 10-Feb-1996.) (Revised by Mario Carneiro, 10-May-2013.)
𝑅 Or 𝑆    &   𝑅 ⊆ (𝑆 × 𝑆)        ¬ 𝐴𝑅𝐴
 
Theoremsotri 5042 A strict order relation is a transitive relation. (Contributed by NM, 10-Feb-1996.) (Revised by Mario Carneiro, 10-May-2013.)
𝑅 Or 𝑆    &   𝑅 ⊆ (𝑆 × 𝑆)       ((𝐴𝑅𝐵𝐵𝑅𝐶) → 𝐴𝑅𝐶)
 
Theoremson2lpi 5043 A strict order relation has no 2-cycle loops. (Contributed by NM, 10-Feb-1996.) (Revised by Mario Carneiro, 10-May-2013.)
𝑅 Or 𝑆    &   𝑅 ⊆ (𝑆 × 𝑆)        ¬ (𝐴𝑅𝐵𝐵𝑅𝐴)
 
Theoremsotri2 5044 A transitivity relation. (Read ¬ B < A and B < C implies A < C .) (Contributed by Mario Carneiro, 10-May-2013.)
𝑅 Or 𝑆    &   𝑅 ⊆ (𝑆 × 𝑆)       ((𝐴𝑆 ∧ ¬ 𝐵𝑅𝐴𝐵𝑅𝐶) → 𝐴𝑅𝐶)
 
Theoremsotri3 5045 A transitivity relation. (Read A < B and ¬ C < B implies A < C .) (Contributed by Mario Carneiro, 10-May-2013.)
𝑅 Or 𝑆    &   𝑅 ⊆ (𝑆 × 𝑆)       ((𝐶𝑆𝐴𝑅𝐵 ∧ ¬ 𝐶𝑅𝐵) → 𝐴𝑅𝐶)
 
Theorempoleloe 5046 Express "less than or equals" for general strict orders. (Contributed by Stefan O'Rear, 17-Jan-2015.)
(𝐵𝑉 → (𝐴(𝑅 ∪ I )𝐵 ↔ (𝐴𝑅𝐵𝐴 = 𝐵)))
 
Theorempoltletr 5047 Transitive law for general strict orders. (Contributed by Stefan O'Rear, 17-Jan-2015.)
((𝑅 Po 𝑋 ∧ (𝐴𝑋𝐵𝑋𝐶𝑋)) → ((𝐴𝑅𝐵𝐵(𝑅 ∪ I )𝐶) → 𝐴𝑅𝐶))
 
Theoremcnvopab 5048* The converse of a class abstraction of ordered pairs. (Contributed by NM, 11-Dec-2003.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
{⟨𝑥, 𝑦⟩ ∣ 𝜑} = {⟨𝑦, 𝑥⟩ ∣ 𝜑}
 
Theoremmptcnv 5049* The converse of a mapping function. (Contributed by Thierry Arnoux, 16-Jan-2017.)
(𝜑 → ((𝑥𝐴𝑦 = 𝐵) ↔ (𝑦𝐶𝑥 = 𝐷)))       (𝜑(𝑥𝐴𝐵) = (𝑦𝐶𝐷))
 
Theoremcnv0 5050 The converse of the empty set. (Contributed by NM, 6-Apr-1998.)
∅ = ∅
 
Theoremcnvi 5051 The converse of the identity relation. Theorem 3.7(ii) of [Monk1] p. 36. (Contributed by NM, 26-Apr-1998.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
I = I
 
Theoremcnvun 5052 The converse of a union is the union of converses. Theorem 16 of [Suppes] p. 62. (Contributed by NM, 25-Mar-1998.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
(𝐴𝐵) = (𝐴𝐵)
 
Theoremcnvdif 5053 Distributive law for converse over set difference. (Contributed by Mario Carneiro, 26-Jun-2014.)
(𝐴𝐵) = (𝐴𝐵)
 
Theoremcnvin 5054 Distributive law for converse over intersection. Theorem 15 of [Suppes] p. 62. (Contributed by NM, 25-Mar-1998.) (Revised by Mario Carneiro, 26-Jun-2014.)
(𝐴𝐵) = (𝐴𝐵)
 
Theoremrnun 5055 Distributive law for range over union. Theorem 8 of [Suppes] p. 60. (Contributed by NM, 24-Mar-1998.)
ran (𝐴𝐵) = (ran 𝐴 ∪ ran 𝐵)
 
Theoremrnin 5056 The range of an intersection belongs the intersection of ranges. Theorem 9 of [Suppes] p. 60. (Contributed by NM, 15-Sep-2004.)
ran (𝐴𝐵) ⊆ (ran 𝐴 ∩ ran 𝐵)
 
Theoremrniun 5057 The range of an indexed union. (Contributed by Mario Carneiro, 29-May-2015.)
ran 𝑥𝐴 𝐵 = 𝑥𝐴 ran 𝐵
 
Theoremrnuni 5058* The range of a union. Part of Exercise 8 of [Enderton] p. 41. (Contributed by NM, 17-Mar-2004.) (Revised by Mario Carneiro, 29-May-2015.)
ran 𝐴 = 𝑥𝐴 ran 𝑥
 
Theoremimaundi 5059 Distributive law for image over union. Theorem 35 of [Suppes] p. 65. (Contributed by NM, 30-Sep-2002.)
(𝐴 “ (𝐵𝐶)) = ((𝐴𝐵) ∪ (𝐴𝐶))
 
Theoremimaundir 5060 The image of a union. (Contributed by Jeff Hoffman, 17-Feb-2008.)
((𝐴𝐵) “ 𝐶) = ((𝐴𝐶) ∪ (𝐵𝐶))
 
Theoremdminss 5061 An upper bound for intersection with a domain. Theorem 40 of [Suppes] p. 66, who calls it "somewhat surprising". (Contributed by NM, 11-Aug-2004.)
(dom 𝑅𝐴) ⊆ (𝑅 “ (𝑅𝐴))
 
Theoremimainss 5062 An upper bound for intersection with an image. Theorem 41 of [Suppes] p. 66. (Contributed by NM, 11-Aug-2004.)
((𝑅𝐴) ∩ 𝐵) ⊆ (𝑅 “ (𝐴 ∩ (𝑅𝐵)))
 
Theoreminimass 5063 The image of an intersection. (Contributed by Thierry Arnoux, 16-Dec-2017.)
((𝐴𝐵) “ 𝐶) ⊆ ((𝐴𝐶) ∩ (𝐵𝐶))
 
Theoreminimasn 5064 The intersection of the image of singleton. (Contributed by Thierry Arnoux, 16-Dec-2017.)
(𝐶𝑉 → ((𝐴𝐵) “ {𝐶}) = ((𝐴 “ {𝐶}) ∩ (𝐵 “ {𝐶})))
 
Theoremcnvxp 5065 The converse of a cross product. Exercise 11 of [Suppes] p. 67. (Contributed by NM, 14-Aug-1999.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
(𝐴 × 𝐵) = (𝐵 × 𝐴)
 
Theoremxp0 5066 The cross product with the empty set is empty. Part of Theorem 3.13(ii) of [Monk1] p. 37. (Contributed by NM, 12-Apr-2004.)
(𝐴 × ∅) = ∅
 
Theoremxpmlem 5067* The cross product of inhabited classes is inhabited. (Contributed by Jim Kingdon, 11-Dec-2018.)
((∃𝑥 𝑥𝐴 ∧ ∃𝑦 𝑦𝐵) ↔ ∃𝑧 𝑧 ∈ (𝐴 × 𝐵))
 
Theoremxpm 5068* The cross product of inhabited classes is inhabited. (Contributed by Jim Kingdon, 13-Dec-2018.)
((∃𝑥 𝑥𝐴 ∧ ∃𝑦 𝑦𝐵) ↔ ∃𝑧 𝑧 ∈ (𝐴 × 𝐵))
 
Theoremxpeq0r 5069 A cross product is empty if at least one member is empty. (Contributed by Jim Kingdon, 12-Dec-2018.)
((𝐴 = ∅ ∨ 𝐵 = ∅) → (𝐴 × 𝐵) = ∅)
 
Theoremsqxpeq0 5070 A Cartesian square is empty iff its member is empty. (Contributed by Jim Kingdon, 21-Apr-2023.)
((𝐴 × 𝐴) = ∅ ↔ 𝐴 = ∅)
 
Theoremxpdisj1 5071 Cross products with disjoint sets are disjoint. (Contributed by NM, 13-Sep-2004.)
((𝐴𝐵) = ∅ → ((𝐴 × 𝐶) ∩ (𝐵 × 𝐷)) = ∅)
 
Theoremxpdisj2 5072 Cross products with disjoint sets are disjoint. (Contributed by NM, 13-Sep-2004.)
((𝐴𝐵) = ∅ → ((𝐶 × 𝐴) ∩ (𝐷 × 𝐵)) = ∅)
 
Theoremxpsndisj 5073 Cross products with two different singletons are disjoint. (Contributed by NM, 28-Jul-2004.)
(𝐵𝐷 → ((𝐴 × {𝐵}) ∩ (𝐶 × {𝐷})) = ∅)
 
Theoremdjudisj 5074* Disjoint unions with disjoint index sets are disjoint. (Contributed by Stefan O'Rear, 21-Nov-2014.)
((𝐴𝐵) = ∅ → ( 𝑥𝐴 ({𝑥} × 𝐶) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅)
 
Theoremresdisj 5075 A double restriction to disjoint classes is the empty set. (Contributed by NM, 7-Oct-2004.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
((𝐴𝐵) = ∅ → ((𝐶𝐴) ↾ 𝐵) = ∅)
 
Theoremrnxpm 5076* The range of a cross product. Part of Theorem 3.13(x) of [Monk1] p. 37, with nonempty changed to inhabited. (Contributed by Jim Kingdon, 12-Dec-2018.)
(∃𝑥 𝑥𝐴 → ran (𝐴 × 𝐵) = 𝐵)
 
Theoremdmxpss 5077 The domain of a cross product is a subclass of the first factor. (Contributed by NM, 19-Mar-2007.)
dom (𝐴 × 𝐵) ⊆ 𝐴
 
Theoremrnxpss 5078 The range of a cross product is a subclass of the second factor. (Contributed by NM, 16-Jan-2006.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
ran (𝐴 × 𝐵) ⊆ 𝐵
 
Theoremdmxpss2 5079 Upper bound for the domain of a binary relation. (Contributed by BJ, 10-Jul-2022.)
(𝑅 ⊆ (𝐴 × 𝐵) → dom 𝑅𝐴)
 
Theoremrnxpss2 5080 Upper bound for the range of a binary relation. (Contributed by BJ, 10-Jul-2022.)
(𝑅 ⊆ (𝐴 × 𝐵) → ran 𝑅𝐵)
 
Theoremrnxpid 5081 The range of a square cross product. (Contributed by FL, 17-May-2010.)
ran (𝐴 × 𝐴) = 𝐴
 
Theoremssxpbm 5082* A cross-product subclass relationship is equivalent to the relationship for its components. (Contributed by Jim Kingdon, 12-Dec-2018.)
(∃𝑥 𝑥 ∈ (𝐴 × 𝐵) → ((𝐴 × 𝐵) ⊆ (𝐶 × 𝐷) ↔ (𝐴𝐶𝐵𝐷)))
 
Theoremssxp1 5083* Cross product subset cancellation. (Contributed by Jim Kingdon, 14-Dec-2018.)
(∃𝑥 𝑥𝐶 → ((𝐴 × 𝐶) ⊆ (𝐵 × 𝐶) ↔ 𝐴𝐵))
 
Theoremssxp2 5084* Cross product subset cancellation. (Contributed by Jim Kingdon, 14-Dec-2018.)
(∃𝑥 𝑥𝐶 → ((𝐶 × 𝐴) ⊆ (𝐶 × 𝐵) ↔ 𝐴𝐵))
 
Theoremxp11m 5085* The cross product of inhabited classes is one-to-one. (Contributed by Jim Kingdon, 13-Dec-2018.)
((∃𝑥 𝑥𝐴 ∧ ∃𝑦 𝑦𝐵) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) ↔ (𝐴 = 𝐶𝐵 = 𝐷)))
 
Theoremxpcanm 5086* Cancellation law for cross-product. (Contributed by Jim Kingdon, 14-Dec-2018.)
(∃𝑥 𝑥𝐶 → ((𝐶 × 𝐴) = (𝐶 × 𝐵) ↔ 𝐴 = 𝐵))
 
Theoremxpcan2m 5087* Cancellation law for cross-product. (Contributed by Jim Kingdon, 14-Dec-2018.)
(∃𝑥 𝑥𝐶 → ((𝐴 × 𝐶) = (𝐵 × 𝐶) ↔ 𝐴 = 𝐵))
 
Theoremxpexr2m 5088* If a nonempty cross product is a set, so are both of its components. (Contributed by Jim Kingdon, 14-Dec-2018.)
(((𝐴 × 𝐵) ∈ 𝐶 ∧ ∃𝑥 𝑥 ∈ (𝐴 × 𝐵)) → (𝐴 ∈ V ∧ 𝐵 ∈ V))
 
Theoremssrnres 5089 Subset of the range of a restriction. (Contributed by NM, 16-Jan-2006.)
(𝐵 ⊆ ran (𝐶𝐴) ↔ ran (𝐶 ∩ (𝐴 × 𝐵)) = 𝐵)
 
Theoremrninxp 5090* Range of the intersection with a cross product. (Contributed by NM, 17-Jan-2006.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
(ran (𝐶 ∩ (𝐴 × 𝐵)) = 𝐵 ↔ ∀𝑦𝐵𝑥𝐴 𝑥𝐶𝑦)
 
Theoremdminxp 5091* Domain of the intersection with a cross product. (Contributed by NM, 17-Jan-2006.)
(dom (𝐶 ∩ (𝐴 × 𝐵)) = 𝐴 ↔ ∀𝑥𝐴𝑦𝐵 𝑥𝐶𝑦)
 
Theoremimainrect 5092 Image of a relation restricted to a rectangular region. (Contributed by Stefan O'Rear, 19-Feb-2015.)
((𝐺 ∩ (𝐴 × 𝐵)) “ 𝑌) = ((𝐺 “ (𝑌𝐴)) ∩ 𝐵)
 
Theoremxpima1 5093 The image by a cross product. (Contributed by Thierry Arnoux, 16-Dec-2017.)
((𝐴𝐶) = ∅ → ((𝐴 × 𝐵) “ 𝐶) = ∅)
 
Theoremxpima2m 5094* The image by a cross product. (Contributed by Thierry Arnoux, 16-Dec-2017.)
(∃𝑥 𝑥 ∈ (𝐴𝐶) → ((𝐴 × 𝐵) “ 𝐶) = 𝐵)
 
Theoremxpimasn 5095 The image of a singleton by a cross product. (Contributed by Thierry Arnoux, 14-Jan-2018.)
(𝑋𝐴 → ((𝐴 × 𝐵) “ {𝑋}) = 𝐵)
 
Theoremcnvcnv3 5096* The set of all ordered pairs in a class is the same as the double converse. (Contributed by Mario Carneiro, 16-Aug-2015.)
𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝑥𝑅𝑦}
 
Theoremdfrel2 5097 Alternate definition of relation. Exercise 2 of [TakeutiZaring] p. 25. (Contributed by NM, 29-Dec-1996.)
(Rel 𝑅𝑅 = 𝑅)
 
Theoremdfrel4v 5098* A relation can be expressed as the set of ordered pairs in it. (Contributed by Mario Carneiro, 16-Aug-2015.)
(Rel 𝑅𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝑥𝑅𝑦})
 
Theoremcnvcnv 5099 The double converse of a class strips out all elements that are not ordered pairs. (Contributed by NM, 8-Dec-2003.)
𝐴 = (𝐴 ∩ (V × V))
 
Theoremcnvcnv2 5100 The double converse of a class equals its restriction to the universe. (Contributed by NM, 8-Oct-2007.)
𝐴 = (𝐴 ↾ V)
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