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Theorem List for Intuitionistic Logic Explorer - 4901-5000   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theorempoirr2 4901 A partial order relation is irreflexive. (Contributed by Mario Carneiro, 2-Nov-2015.)
(𝑅 Po 𝐴 → (𝑅 ∩ ( I ↾ 𝐴)) = ∅)
 
Theoremtrinxp 4902 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 4903 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 4904 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 4905 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 4906 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 4907 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 4908 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 4909 Transitive law for general strict orders. (Contributed by Stefan O'Rear, 17-Jan-2015.)
((𝑅 Po 𝑋 ∧ (𝐴𝑋𝐵𝑋𝐶𝑋)) → ((𝐴𝑅𝐵𝐵(𝑅 ∪ I )𝐶) → 𝐴𝑅𝐶))
 
Theoremcnvopab 4910* 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 4911* The converse of a mapping function. (Contributed by Thierry Arnoux, 16-Jan-2017.)
(𝜑 → ((𝑥𝐴𝑦 = 𝐵) ↔ (𝑦𝐶𝑥 = 𝐷)))       (𝜑(𝑥𝐴𝐵) = (𝑦𝐶𝐷))
 
Theoremcnv0 4912 The converse of the empty set. (Contributed by NM, 6-Apr-1998.)
∅ = ∅
 
Theoremcnvi 4913 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 4914 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 4915 Distributive law for converse over set difference. (Contributed by Mario Carneiro, 26-Jun-2014.)
(𝐴𝐵) = (𝐴𝐵)
 
Theoremcnvin 4916 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 4917 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 4918 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 4919 The range of an indexed union. (Contributed by Mario Carneiro, 29-May-2015.)
ran 𝑥𝐴 𝐵 = 𝑥𝐴 ran 𝐵
 
Theoremrnuni 4920* 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 4921 Distributive law for image over union. Theorem 35 of [Suppes] p. 65. (Contributed by NM, 30-Sep-2002.)
(𝐴 “ (𝐵𝐶)) = ((𝐴𝐵) ∪ (𝐴𝐶))
 
Theoremimaundir 4922 The image of a union. (Contributed by Jeff Hoffman, 17-Feb-2008.)
((𝐴𝐵) “ 𝐶) = ((𝐴𝐶) ∪ (𝐵𝐶))
 
Theoremdminss 4923 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 4924 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 4925 The image of an intersection (Contributed by Thierry Arnoux, 16-Dec-2017.)
((𝐴𝐵) “ 𝐶) ⊆ ((𝐴𝐶) ∩ (𝐵𝐶))
 
Theoreminimasn 4926 The intersection of the image of singleton (Contributed by Thierry Arnoux, 16-Dec-2017.)
(𝐶𝑉 → ((𝐴𝐵) “ {𝐶}) = ((𝐴 “ {𝐶}) ∩ (𝐵 “ {𝐶})))
 
Theoremcnvxp 4927 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 4928 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 4929* The cross product of inhabited classes is inhabited. (Contributed by Jim Kingdon, 11-Dec-2018.)
((∃𝑥 𝑥𝐴 ∧ ∃𝑦 𝑦𝐵) ↔ ∃𝑧 𝑧 ∈ (𝐴 × 𝐵))
 
Theoremxpm 4930* The cross product of inhabited classes is inhabited. (Contributed by Jim Kingdon, 13-Dec-2018.)
((∃𝑥 𝑥𝐴 ∧ ∃𝑦 𝑦𝐵) ↔ ∃𝑧 𝑧 ∈ (𝐴 × 𝐵))
 
Theoremxpeq0r 4931 A cross product is empty if at least one member is empty. (Contributed by Jim Kingdon, 12-Dec-2018.)
((𝐴 = ∅ ∨ 𝐵 = ∅) → (𝐴 × 𝐵) = ∅)
 
Theoremsqxpeq0 4932 A Cartesian square is empty iff its member is empty. (Contributed by Jim Kingdon, 21-Apr-2023.)
((𝐴 × 𝐴) = ∅ ↔ 𝐴 = ∅)
 
Theoremxpdisj1 4933 Cross products with disjoint sets are disjoint. (Contributed by NM, 13-Sep-2004.)
((𝐴𝐵) = ∅ → ((𝐴 × 𝐶) ∩ (𝐵 × 𝐷)) = ∅)
 
Theoremxpdisj2 4934 Cross products with disjoint sets are disjoint. (Contributed by NM, 13-Sep-2004.)
((𝐴𝐵) = ∅ → ((𝐶 × 𝐴) ∩ (𝐷 × 𝐵)) = ∅)
 
Theoremxpsndisj 4935 Cross products with two different singletons are disjoint. (Contributed by NM, 28-Jul-2004.)
(𝐵𝐷 → ((𝐴 × {𝐵}) ∩ (𝐶 × {𝐷})) = ∅)
 
Theoremdjudisj 4936* Disjoint unions with disjoint index sets are disjoint. (Contributed by Stefan O'Rear, 21-Nov-2014.)
((𝐴𝐵) = ∅ → ( 𝑥𝐴 ({𝑥} × 𝐶) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅)
 
Theoremresdisj 4937 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 4938* 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 4939 The domain of a cross product is a subclass of the first factor. (Contributed by NM, 19-Mar-2007.)
dom (𝐴 × 𝐵) ⊆ 𝐴
 
Theoremrnxpss 4940 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 4941 Upper bound for the domain of a binary relation. (Contributed by BJ, 10-Jul-2022.)
(𝑅 ⊆ (𝐴 × 𝐵) → dom 𝑅𝐴)
 
Theoremrnxpss2 4942 Upper bound for the range of a binary relation. (Contributed by BJ, 10-Jul-2022.)
(𝑅 ⊆ (𝐴 × 𝐵) → ran 𝑅𝐵)
 
Theoremrnxpid 4943 The range of a square cross product. (Contributed by FL, 17-May-2010.)
ran (𝐴 × 𝐴) = 𝐴
 
Theoremssxpbm 4944* 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 4945* Cross product subset cancellation. (Contributed by Jim Kingdon, 14-Dec-2018.)
(∃𝑥 𝑥𝐶 → ((𝐴 × 𝐶) ⊆ (𝐵 × 𝐶) ↔ 𝐴𝐵))
 
Theoremssxp2 4946* Cross product subset cancellation. (Contributed by Jim Kingdon, 14-Dec-2018.)
(∃𝑥 𝑥𝐶 → ((𝐶 × 𝐴) ⊆ (𝐶 × 𝐵) ↔ 𝐴𝐵))
 
Theoremxp11m 4947* 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 4948* Cancellation law for cross-product. (Contributed by Jim Kingdon, 14-Dec-2018.)
(∃𝑥 𝑥𝐶 → ((𝐶 × 𝐴) = (𝐶 × 𝐵) ↔ 𝐴 = 𝐵))
 
Theoremxpcan2m 4949* Cancellation law for cross-product. (Contributed by Jim Kingdon, 14-Dec-2018.)
(∃𝑥 𝑥𝐶 → ((𝐴 × 𝐶) = (𝐵 × 𝐶) ↔ 𝐴 = 𝐵))
 
Theoremxpexr2m 4950* 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 4951 Subset of the range of a restriction. (Contributed by NM, 16-Jan-2006.)
(𝐵 ⊆ ran (𝐶𝐴) ↔ ran (𝐶 ∩ (𝐴 × 𝐵)) = 𝐵)
 
Theoremrninxp 4952* 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 4953* Domain of the intersection with a cross product. (Contributed by NM, 17-Jan-2006.)
(dom (𝐶 ∩ (𝐴 × 𝐵)) = 𝐴 ↔ ∀𝑥𝐴𝑦𝐵 𝑥𝐶𝑦)
 
Theoremimainrect 4954 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 4955 The image by a cross product. (Contributed by Thierry Arnoux, 16-Dec-2017.)
((𝐴𝐶) = ∅ → ((𝐴 × 𝐵) “ 𝐶) = ∅)
 
Theoremxpima2m 4956* The image by a cross product. (Contributed by Thierry Arnoux, 16-Dec-2017.)
(∃𝑥 𝑥 ∈ (𝐴𝐶) → ((𝐴 × 𝐵) “ 𝐶) = 𝐵)
 
Theoremxpimasn 4957 The image of a singleton by a cross product. (Contributed by Thierry Arnoux, 14-Jan-2018.)
(𝑋𝐴 → ((𝐴 × 𝐵) “ {𝑋}) = 𝐵)
 
Theoremcnvcnv3 4958* 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 4959 Alternate definition of relation. Exercise 2 of [TakeutiZaring] p. 25. (Contributed by NM, 29-Dec-1996.)
(Rel 𝑅𝑅 = 𝑅)
 
Theoremdfrel4v 4960* 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 4961 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 4962 The double converse of a class equals its restriction to the universe. (Contributed by NM, 8-Oct-2007.)
𝐴 = (𝐴 ↾ V)
 
Theoremcnvcnvss 4963 The double converse of a class is a subclass. Exercise 2 of [TakeutiZaring] p. 25. (Contributed by NM, 23-Jul-2004.)
𝐴𝐴
 
Theoremcnveqb 4964 Equality theorem for converse. (Contributed by FL, 19-Sep-2011.)
((Rel 𝐴 ∧ Rel 𝐵) → (𝐴 = 𝐵𝐴 = 𝐵))
 
Theoremcnveq0 4965 A relation empty iff its converse is empty. (Contributed by FL, 19-Sep-2011.)
(Rel 𝐴 → (𝐴 = ∅ ↔ 𝐴 = ∅))
 
Theoremdfrel3 4966 Alternate definition of relation. (Contributed by NM, 14-May-2008.)
(Rel 𝑅 ↔ (𝑅 ↾ V) = 𝑅)
 
Theoremdmresv 4967 The domain of a universal restriction. (Contributed by NM, 14-May-2008.)
dom (𝐴 ↾ V) = dom 𝐴
 
Theoremrnresv 4968 The range of a universal restriction. (Contributed by NM, 14-May-2008.)
ran (𝐴 ↾ V) = ran 𝐴
 
Theoremdfrn4 4969 Range defined in terms of image. (Contributed by NM, 14-May-2008.)
ran 𝐴 = (𝐴 “ V)
 
Theoremcsbrng 4970 Distribute proper substitution through the range of a class. (Contributed by Alan Sare, 10-Nov-2012.)
(𝐴𝑉𝐴 / 𝑥ran 𝐵 = ran 𝐴 / 𝑥𝐵)
 
Theoremrescnvcnv 4971 The restriction of the double converse of a class. (Contributed by NM, 8-Apr-2007.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
(𝐴𝐵) = (𝐴𝐵)
 
Theoremcnvcnvres 4972 The double converse of the restriction of a class. (Contributed by NM, 3-Jun-2007.)
(𝐴𝐵) = (𝐴𝐵)
 
Theoremimacnvcnv 4973 The image of the double converse of a class. (Contributed by NM, 8-Apr-2007.)
(𝐴𝐵) = (𝐴𝐵)
 
Theoremdmsnm 4974* The domain of a singleton is inhabited iff the singleton argument is an ordered pair. (Contributed by Jim Kingdon, 15-Dec-2018.)
(𝐴 ∈ (V × V) ↔ ∃𝑥 𝑥 ∈ dom {𝐴})
 
Theoremrnsnm 4975* The range of a singleton is inhabited iff the singleton argument is an ordered pair. (Contributed by Jim Kingdon, 15-Dec-2018.)
(𝐴 ∈ (V × V) ↔ ∃𝑥 𝑥 ∈ ran {𝐴})
 
Theoremdmsn0 4976 The domain of the singleton of the empty set is empty. (Contributed by NM, 30-Jan-2004.)
dom {∅} = ∅
 
Theoremcnvsn0 4977 The converse of the singleton of the empty set is empty. (Contributed by Mario Carneiro, 30-Aug-2015.)
{∅} = ∅
 
Theoremdmsn0el 4978 The domain of a singleton is empty if the singleton's argument contains the empty set. (Contributed by NM, 15-Dec-2008.)
(∅ ∈ 𝐴 → dom {𝐴} = ∅)
 
Theoremrelsn2m 4979* A singleton is a relation iff it has an inhabited domain. (Contributed by Jim Kingdon, 16-Dec-2018.)
𝐴 ∈ V       (Rel {𝐴} ↔ ∃𝑥 𝑥 ∈ dom {𝐴})
 
Theoremdmsnopg 4980 The domain of a singleton of an ordered pair is the singleton of the first member. (Contributed by Mario Carneiro, 26-Apr-2015.)
(𝐵𝑉 → dom {⟨𝐴, 𝐵⟩} = {𝐴})
 
Theoremdmpropg 4981 The domain of an unordered pair of ordered pairs. (Contributed by Mario Carneiro, 26-Apr-2015.)
((𝐵𝑉𝐷𝑊) → dom {⟨𝐴, 𝐵⟩, ⟨𝐶, 𝐷⟩} = {𝐴, 𝐶})
 
Theoremdmsnop 4982 The domain of a singleton of an ordered pair is the singleton of the first member. (Contributed by NM, 30-Jan-2004.) (Proof shortened by Andrew Salmon, 27-Aug-2011.) (Revised by Mario Carneiro, 26-Apr-2015.)
𝐵 ∈ V       dom {⟨𝐴, 𝐵⟩} = {𝐴}
 
Theoremdmprop 4983 The domain of an unordered pair of ordered pairs. (Contributed by NM, 13-Sep-2011.)
𝐵 ∈ V    &   𝐷 ∈ V       dom {⟨𝐴, 𝐵⟩, ⟨𝐶, 𝐷⟩} = {𝐴, 𝐶}
 
Theoremdmtpop 4984 The domain of an unordered triple of ordered pairs. (Contributed by NM, 14-Sep-2011.)
𝐵 ∈ V    &   𝐷 ∈ V    &   𝐹 ∈ V       dom {⟨𝐴, 𝐵⟩, ⟨𝐶, 𝐷⟩, ⟨𝐸, 𝐹⟩} = {𝐴, 𝐶, 𝐸}
 
Theoremcnvcnvsn 4985 Double converse of a singleton of an ordered pair. (Unlike cnvsn 4991, this does not need any sethood assumptions on 𝐴 and 𝐵.) (Contributed by Mario Carneiro, 26-Apr-2015.)
{⟨𝐴, 𝐵⟩} = {⟨𝐵, 𝐴⟩}
 
Theoremdmsnsnsng 4986 The domain of the singleton of the singleton of a singleton. (Contributed by Jim Kingdon, 16-Dec-2018.)
(𝐴 ∈ V → dom {{{𝐴}}} = {𝐴})
 
Theoremrnsnopg 4987 The range of a singleton of an ordered pair is the singleton of the second member. (Contributed by NM, 24-Jul-2004.) (Revised by Mario Carneiro, 30-Apr-2015.)
(𝐴𝑉 → ran {⟨𝐴, 𝐵⟩} = {𝐵})
 
Theoremrnpropg 4988 The range of a pair of ordered pairs is the pair of second members. (Contributed by Thierry Arnoux, 3-Jan-2017.)
((𝐴𝑉𝐵𝑊) → ran {⟨𝐴, 𝐶⟩, ⟨𝐵, 𝐷⟩} = {𝐶, 𝐷})
 
Theoremrnsnop 4989 The range of a singleton of an ordered pair is the singleton of the second member. (Contributed by NM, 24-Jul-2004.) (Revised by Mario Carneiro, 26-Apr-2015.)
𝐴 ∈ V       ran {⟨𝐴, 𝐵⟩} = {𝐵}
 
Theoremop1sta 4990 Extract the first member of an ordered pair. (See op2nda 4993 to extract the second member and op1stb 4369 for an alternate version.) (Contributed by Raph Levien, 4-Dec-2003.)
𝐴 ∈ V    &   𝐵 ∈ V        dom {⟨𝐴, 𝐵⟩} = 𝐴
 
Theoremcnvsn 4991 Converse of a singleton of an ordered pair. (Contributed by NM, 11-May-1998.) (Revised by Mario Carneiro, 26-Apr-2015.)
𝐴 ∈ V    &   𝐵 ∈ V       {⟨𝐴, 𝐵⟩} = {⟨𝐵, 𝐴⟩}
 
Theoremop2ndb 4992 Extract the second member of an ordered pair. Theorem 5.12(ii) of [Monk1] p. 52. (See op1stb 4369 to extract the first member and op2nda 4993 for an alternate version.) (Contributed by NM, 25-Nov-2003.)
𝐴 ∈ V    &   𝐵 ∈ V        {⟨𝐴, 𝐵⟩} = 𝐵
 
Theoremop2nda 4993 Extract the second member of an ordered pair. (See op1sta 4990 to extract the first member and op2ndb 4992 for an alternate version.) (Contributed by NM, 17-Feb-2004.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
𝐴 ∈ V    &   𝐵 ∈ V        ran {⟨𝐴, 𝐵⟩} = 𝐵
 
Theoremcnvsng 4994 Converse of a singleton of an ordered pair. (Contributed by NM, 23-Jan-2015.)
((𝐴𝑉𝐵𝑊) → {⟨𝐴, 𝐵⟩} = {⟨𝐵, 𝐴⟩})
 
Theoremopswapg 4995 Swap the members of an ordered pair. (Contributed by Jim Kingdon, 16-Dec-2018.)
((𝐴𝑉𝐵𝑊) → {⟨𝐴, 𝐵⟩} = ⟨𝐵, 𝐴⟩)
 
Theoremelxp4 4996 Membership in a cross product. This version requires no quantifiers or dummy variables. See also elxp5 4997. (Contributed by NM, 17-Feb-2004.)
(𝐴 ∈ (𝐵 × 𝐶) ↔ (𝐴 = ⟨ dom {𝐴}, ran {𝐴}⟩ ∧ ( dom {𝐴} ∈ 𝐵 ran {𝐴} ∈ 𝐶)))
 
Theoremelxp5 4997 Membership in a cross product requiring no quantifiers or dummy variables. Provides a slightly shorter version of elxp4 4996 when the double intersection does not create class existence problems (caused by int0 3755). (Contributed by NM, 1-Aug-2004.)
(𝐴 ∈ (𝐵 × 𝐶) ↔ (𝐴 = ⟨ 𝐴, ran {𝐴}⟩ ∧ ( 𝐴𝐵 ran {𝐴} ∈ 𝐶)))
 
Theoremcnvresima 4998 An image under the converse of a restriction. (Contributed by Jeff Hankins, 12-Jul-2009.)
((𝐹𝐴) “ 𝐵) = ((𝐹𝐵) ∩ 𝐴)
 
Theoremresdm2 4999 A class restricted to its domain equals its double converse. (Contributed by NM, 8-Apr-2007.)
(𝐴 ↾ dom 𝐴) = 𝐴
 
Theoremresdmres 5000 Restriction to the domain of a restriction. (Contributed by NM, 8-Apr-2007.)
(𝐴 ↾ dom (𝐴𝐵)) = (𝐴𝐵)
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