P. 62 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 6101-6200   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	rnuni 6101*	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 = ∪ 𝑥 ∈ 𝐴 ran 𝑥
Theorem	imaundi 6102	Distributive law for image over union. Theorem 35 of [Suppes] p. 65. (Contributed by NM, 30-Sep-2002.)
⊢ (𝐴 “ (𝐵 ∪ 𝐶)) = ((𝐴 “ 𝐵) ∪ (𝐴 “ 𝐶))
Theorem	imaundir 6103	The image of a union. (Contributed by Jeff Hoffman, 17-Feb-2008.)
⊢ ((𝐴 ∪ 𝐵) “ 𝐶) = ((𝐴 “ 𝐶) ∪ (𝐵 “ 𝐶))
Theorem	imadifssran 6104	Condition for the range of a relation to be the range of one its restrictions. (Contributed by AV, 4-Oct-2025.)
⊢ ((Rel 𝐹 ∧ 𝐴 ⊆ dom 𝐹) → ((𝐹 “ (dom 𝐹 ∖ 𝐴)) ⊆ ran (𝐹 ↾ 𝐴) → ran 𝐹 = ran (𝐹 ↾ 𝐴)))
Theorem	cnvimassrndm 6105	The preimage of a superset of the range of a class is the domain of the class. Generalization of cnvimarndm 6038 for subsets. (Contributed by AV, 18-Sep-2024.)
⊢ (ran 𝐹 ⊆ 𝐴 → (◡𝐹 “ 𝐴) = dom 𝐹)
Theorem	dminss 6106	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 𝑅 ∩ 𝐴) ⊆ (◡𝑅 “ (𝑅 “ 𝐴))
Theorem	imainss 6107	An upper bound for intersection with an image. Theorem 41 of [Suppes] p. 66. (Contributed by NM, 11-Aug-2004.)
⊢ ((𝑅 “ 𝐴) ∩ 𝐵) ⊆ (𝑅 “ (𝐴 ∩ (◡𝑅 “ 𝐵)))
Theorem	inimass 6108	The image of an intersection. (Contributed by Thierry Arnoux, 16-Dec-2017.)
⊢ ((𝐴 ∩ 𝐵) “ 𝐶) ⊆ ((𝐴 “ 𝐶) ∩ (𝐵 “ 𝐶))
Theorem	inimasn 6109	The intersection of the image of singleton. (Contributed by Thierry Arnoux, 16-Dec-2017.)
⊢ (𝐶 ∈ 𝑉 → ((𝐴 ∩ 𝐵) “ {𝐶}) = ((𝐴 “ {𝐶}) ∩ (𝐵 “ {𝐶})))
Theorem	cnvxp 6110	The converse of a Cartesian 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)
Theorem	xp0 6111	The Cartesian product with the empty set is empty. Part of Theorem 3.13(ii) of [Monk1] p. 37. (Contributed by NM, 12-Apr-2004.)
⊢ (𝐴 × ∅) = ∅
Theorem	xpnz 6112	The Cartesian product of nonempty classes is nonempty. (Variation of a theorem contributed by Raph Levien, 30-Jun-2006.) (Contributed by NM, 30-Jun-2006.)
⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) ↔ (𝐴 × 𝐵) ≠ ∅)
Theorem	xpeq0 6113	At least one member of an empty Cartesian product is empty. (Contributed by NM, 27-Aug-2006.)
⊢ ((𝐴 × 𝐵) = ∅ ↔ (𝐴 = ∅ ∨ 𝐵 = ∅))
Theorem	xpdisj1 6114	Cartesian products with disjoint sets are disjoint. (Contributed by NM, 13-Sep-2004.)
⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 × 𝐶) ∩ (𝐵 × 𝐷)) = ∅)
Theorem	xpdisj2 6115	Cartesian products with disjoint sets are disjoint. (Contributed by NM, 13-Sep-2004.)
⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐶 × 𝐴) ∩ (𝐷 × 𝐵)) = ∅)
Theorem	xpsndisj 6116	Cartesian products with two different singletons are disjoint. (Contributed by NM, 28-Jul-2004.)
⊢ (𝐵 ≠ 𝐷 → ((𝐴 × {𝐵}) ∩ (𝐶 × {𝐷})) = ∅)
Theorem	difxp 6117	Difference of Cartesian products, expressed in terms of a union of Cartesian products of differences. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 26-Jun-2014.) (Proof shortened by Wolf Lammen, 16-May-2025.)
⊢ ((𝐶 × 𝐷) ∖ (𝐴 × 𝐵)) = (((𝐶 ∖ 𝐴) × 𝐷) ∪ (𝐶 × (𝐷 ∖ 𝐵)))
Theorem	difxp1 6118	Difference law for Cartesian product. (Contributed by Scott Fenton, 18-Feb-2013.) (Revised by Mario Carneiro, 26-Jun-2014.)
⊢ ((𝐴 ∖ 𝐵) × 𝐶) = ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶))
Theorem	difxp2 6119	Difference law for Cartesian product. (Contributed by Scott Fenton, 18-Feb-2013.) (Revised by Mario Carneiro, 26-Jun-2014.)
⊢ (𝐴 × (𝐵 ∖ 𝐶)) = ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶))
Theorem	djudisj 6120*	Disjoint unions with disjoint index sets are disjoint. (Contributed by Stefan O'Rear, 21-Nov-2014.)
⊢ ((𝐴 ∩ 𝐵) = ∅ → (∪ 𝑥 ∈ 𝐴 ({𝑥} × 𝐶) ∩ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷)) = ∅)
Theorem	xpdifid 6121*	The set of distinct couples in a Cartesian product. (Contributed by Thierry Arnoux, 25-May-2019.)
⊢ ∪ 𝑥 ∈ 𝐴 ({𝑥} × (𝐵 ∖ {𝑥})) = ((𝐴 × 𝐵) ∖ I )
Theorem	resdisj 6122	A double restriction to disjoint classes is the empty set. (Contributed by NM, 7-Oct-2004.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐶 ↾ 𝐴) ↾ 𝐵) = ∅)
Theorem	rnxp 6123	The range of a Cartesian product. Part of Theorem 3.13(x) of [Monk1] p. 37. (Contributed by NM, 12-Apr-2004.)
⊢ (𝐴 ≠ ∅ → ran (𝐴 × 𝐵) = 𝐵)
Theorem	dmxpss 6124	The domain of a Cartesian product is included in its first factor. (Contributed by NM, 19-Mar-2007.)
⊢ dom (𝐴 × 𝐵) ⊆ 𝐴
Theorem	rnxpss 6125	The range of a Cartesian product is included in its second factor. (Contributed by NM, 16-Jan-2006.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
⊢ ran (𝐴 × 𝐵) ⊆ 𝐵
Theorem	rnxpid 6126	The range of a Cartesian square. (Contributed by FL, 17-May-2010.)
⊢ ran (𝐴 × 𝐴) = 𝐴
Theorem	ssxpb 6127	A Cartesian product subclass relationship is equivalent to the conjunction of the analogous relationships for the factors. (Contributed by NM, 17-Dec-2008.)
⊢ ((𝐴 × 𝐵) ≠ ∅ → ((𝐴 × 𝐵) ⊆ (𝐶 × 𝐷) ↔ (𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷)))
Theorem	xp11 6128	The Cartesian product of nonempty classes is a one-to-one "function" of its two "arguments". In other words, two Cartesian products, at least one with nonempty factors, are equal if and only if their respective factors are equal. (Contributed by NM, 31-May-2008.)
⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
Theorem	xpcan 6129	Cancellation law for Cartesian product. (Contributed by NM, 30-Aug-2011.)
⊢ (𝐶 ≠ ∅ → ((𝐶 × 𝐴) = (𝐶 × 𝐵) ↔ 𝐴 = 𝐵))
Theorem	xpcan2 6130	Cancellation law for Cartesian product. (Contributed by NM, 30-Aug-2011.)
⊢ (𝐶 ≠ ∅ → ((𝐴 × 𝐶) = (𝐵 × 𝐶) ↔ 𝐴 = 𝐵))
Theorem	ssrnres 6131	Two ways to express surjectivity of a restricted and corestricted binary relation (intersection of a binary relation with a Cartesian product): the LHS expresses inclusion in the range of the restricted relation, while the RHS expresses equality with the range of the restricted and corestricted relation. (Contributed by NM, 16-Jan-2006.) (Proof shortened by Peter Mazsa, 2-Oct-2022.)
⊢ (𝐵 ⊆ ran (𝐶 ↾ 𝐴) ↔ ran (𝐶 ∩ (𝐴 × 𝐵)) = 𝐵)
Theorem	rninxp 6132*	Two ways to express surjectivity of a restricted and corestricted binary relation (intersection of a binary relation with a Cartesian product). (Contributed by NM, 17-Jan-2006.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
⊢ (ran (𝐶 ∩ (𝐴 × 𝐵)) = 𝐵 ↔ ∀𝑦 ∈ 𝐵 ∃𝑥 ∈ 𝐴 𝑥𝐶𝑦)
Theorem	dminxp 6133*	Two ways to express totality of a restricted and corestricted binary relation (intersection of a binary relation with a Cartesian product). (Contributed by NM, 17-Jan-2006.)
⊢ (dom (𝐶 ∩ (𝐴 × 𝐵)) = 𝐴 ↔ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑥𝐶𝑦)
Theorem	imainrect 6134	Image by a restricted and corestricted binary relation (intersection of a binary relation with a Cartesian product). (Contributed by Stefan O'Rear, 19-Feb-2015.)
⊢ ((𝐺 ∩ (𝐴 × 𝐵)) “ 𝑌) = ((𝐺 “ (𝑌 ∩ 𝐴)) ∩ 𝐵)
Theorem	xpima 6135	Direct image by a Cartesian product. (Contributed by Thierry Arnoux, 4-Feb-2017.)
⊢ ((𝐴 × 𝐵) “ 𝐶) = if((𝐴 ∩ 𝐶) = ∅, ∅, 𝐵)
Theorem	xpima1 6136	Direct image by a Cartesian product (case of empty intersection with the domain). (Contributed by Thierry Arnoux, 16-Dec-2017.)
⊢ ((𝐴 ∩ 𝐶) = ∅ → ((𝐴 × 𝐵) “ 𝐶) = ∅)
Theorem	xpima2 6137	Direct image by a Cartesian product (case of nonempty intersection with the domain). (Contributed by Thierry Arnoux, 16-Dec-2017.)
⊢ ((𝐴 ∩ 𝐶) ≠ ∅ → ((𝐴 × 𝐵) “ 𝐶) = 𝐵)
Theorem	xpimasn 6138	Direct image of a singleton by a Cartesian product. (Contributed by Thierry Arnoux, 14-Jan-2018.) (Proof shortened by BJ, 6-Apr-2019.)
⊢ (𝑋 ∈ 𝐴 → ((𝐴 × 𝐵) “ {𝑋}) = 𝐵)
Theorem	sossfld 6139	The base set of a strict order is contained in the field of the relation, except possibly for one element (note that ∅ Or {𝐵}). (Contributed by Mario Carneiro, 27-Apr-2015.)
⊢ ((𝑅 Or 𝐴 ∧ 𝐵 ∈ 𝐴) → (𝐴 ∖ {𝐵}) ⊆ (dom 𝑅 ∪ ran 𝑅))
Theorem	sofld 6140	The base set of a nonempty strict order is the same as the field of the relation. (Contributed by Mario Carneiro, 15-May-2015.)
⊢ ((𝑅 Or 𝐴 ∧ 𝑅 ⊆ (𝐴 × 𝐴) ∧ 𝑅 ≠ ∅) → 𝐴 = (dom 𝑅 ∪ ran 𝑅))
Theorem	cnvcnv3 6141*	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}
Theorem	dfrel2 6142	Alternate definition of relation. Exercise 2 of [TakeutiZaring] p. 25. (Contributed by NM, 29-Dec-1996.)
⊢ (Rel 𝑅 ↔ ◡◡𝑅 = 𝑅)
Theorem	dfrel4v 6143*	A relation can be expressed as the set of ordered pairs in it. An analogue of dffn5 6885 for relations. (Contributed by Mario Carneiro, 16-Aug-2015.)
⊢ (Rel 𝑅 ↔ 𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝑥𝑅𝑦})
Theorem	dfrel4 6144*	A relation can be expressed as the set of ordered pairs in it. An analogue of dffn5 6885 for relations. (Contributed by Mario Carneiro, 16-Aug-2015.) (Revised by Thierry Arnoux, 11-May-2017.)
⊢ Ⅎ𝑥𝑅    &   ⊢ Ⅎ𝑦𝑅    ⇒   ⊢ (Rel 𝑅 ↔ 𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝑥𝑅𝑦})
Theorem	cnvcnv 6145	The double converse of a class strips out all elements that are not ordered pairs. (Contributed by NM, 8-Dec-2003.) (Proof shortened by BJ, 26-Nov-2021.)
⊢ ◡◡𝐴 = (𝐴 ∩ (V × V))
Theorem	cnvcnv2 6146	The double converse of a class equals its restriction to the universe. (Contributed by NM, 8-Oct-2007.)
⊢ ◡◡𝐴 = (𝐴 ↾ V)
Theorem	cnvcnvss 6147	The double converse of a class is a subclass. Exercise 2 of [TakeutiZaring] p. 25. (Contributed by NM, 23-Jul-2004.)
⊢ ◡◡𝐴 ⊆ 𝐴
Theorem	cnvrescnv 6148	Two ways to express the corestriction of a class. (Contributed by BJ, 28-Dec-2023.)
⊢ ◡(◡𝑅 ↾ 𝐵) = (𝑅 ∩ (V × 𝐵))
Theorem	cnveqb 6149	Equality theorem for converse. (Contributed by FL, 19-Sep-2011.)
⊢ ((Rel 𝐴 ∧ Rel 𝐵) → (𝐴 = 𝐵 ↔ ◡𝐴 = ◡𝐵))
Theorem	cnveq0 6150	A relation empty iff its converse is empty. (Contributed by FL, 19-Sep-2011.)
⊢ (Rel 𝐴 → (𝐴 = ∅ ↔ ◡𝐴 = ∅))
Theorem	dfrel3 6151	Alternate definition of relation. (Contributed by NM, 14-May-2008.)
⊢ (Rel 𝑅 ↔ (𝑅 ↾ V) = 𝑅)
Theorem	elid 6152*	Characterization of the elements of the identity relation. TODO: reorder theorems to move this theorem and dfrel3 6151 after elrid 6001. (Contributed by BJ, 28-Aug-2022.)
⊢ (𝐴 ∈ I ↔ ∃𝑥 𝐴 = ⟨𝑥, 𝑥⟩)
Theorem	dmresv 6153	The domain of a universal restriction. (Contributed by NM, 14-May-2008.)
⊢ dom (𝐴 ↾ V) = dom 𝐴
Theorem	rnresv 6154	The range of a universal restriction. (Contributed by NM, 14-May-2008.)
⊢ ran (𝐴 ↾ V) = ran 𝐴
Theorem	dfrn4 6155	Range defined in terms of image. (Contributed by NM, 14-May-2008.)
⊢ ran 𝐴 = (𝐴 “ V)
Theorem	csbrn 6156	Distribute proper substitution through the range of a class. (Contributed by Alan Sare, 10-Nov-2012.)
⊢ ⦋𝐴 / 𝑥⦌ran 𝐵 = ran ⦋𝐴 / 𝑥⦌𝐵
Theorem	rescnvcnv 6157	The restriction of the double converse of a class. (Contributed by NM, 8-Apr-2007.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
⊢ (◡◡𝐴 ↾ 𝐵) = (𝐴 ↾ 𝐵)
Theorem	cnvcnvres 6158	The double converse of the restriction of a class. (Contributed by NM, 3-Jun-2007.)
⊢ ◡◡(𝐴 ↾ 𝐵) = (◡◡𝐴 ↾ 𝐵)
Theorem	imacnvcnv 6159	The image of the double converse of a class. (Contributed by NM, 8-Apr-2007.)
⊢ (◡◡𝐴 “ 𝐵) = (𝐴 “ 𝐵)
Theorem	dmsnn0 6160	The domain of a singleton is nonzero iff the singleton argument is an ordered pair. (Contributed by NM, 14-Dec-2008.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
⊢ (𝐴 ∈ (V × V) ↔ dom {𝐴} ≠ ∅)
Theorem	rnsnn0 6161	The range of a singleton is nonzero iff the singleton argument is an ordered pair. (Contributed by NM, 14-Dec-2008.)
⊢ (𝐴 ∈ (V × V) ↔ ran {𝐴} ≠ ∅)
Theorem	dmsn0 6162	The domain of the singleton of the empty set is empty. (Contributed by NM, 30-Jan-2004.)
⊢ dom {∅} = ∅
Theorem	cnvsn0 6163	The converse of the singleton of the empty set is empty. (Contributed by Mario Carneiro, 30-Aug-2015.)
⊢ ◡{∅} = ∅
Theorem	dmsn0el 6164	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} = ∅)
Theorem	relsn2 6165	A singleton is a relation iff it has a nonempty domain. (Contributed by NM, 25-Sep-2013.) Make hypothesis an antecedent. (Revised by BJ, 12-Feb-2022.)
⊢ (𝐴 ∈ 𝑉 → (Rel {𝐴} ↔ dom {𝐴} ≠ ∅))
Theorem	dmsnopg 6166	The domain of a singleton of an ordered pair is the singleton of the first member. (Contributed by Mario Carneiro, 26-Apr-2015.)
⊢ (𝐵 ∈ 𝑉 → dom {⟨𝐴, 𝐵⟩} = {𝐴})
Theorem	dmsnopss 6167	The domain of a singleton of an ordered pair is a subset of the singleton of the first member (with no sethood assumptions on 𝐵). (Contributed by Mario Carneiro, 30-Apr-2015.)
⊢ dom {⟨𝐴, 𝐵⟩} ⊆ {𝐴}
Theorem	dmpropg 6168	The domain of an unordered pair of ordered pairs. (Contributed by Mario Carneiro, 26-Apr-2015.)
⊢ ((𝐵 ∈ 𝑉 ∧ 𝐷 ∈ 𝑊) → dom {⟨𝐴, 𝐵⟩, ⟨𝐶, 𝐷⟩} = {𝐴, 𝐶})
Theorem	dmsnop 6169	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}
Theorem	dmprop 6170	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}
Theorem	dmtpop 6171	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}
Theorem	cnvcnvsn 6172	Double converse of a singleton of an ordered pair. (Unlike cnvsn 6179, this does not need any sethood assumptions on 𝐴 and 𝐵.) (Contributed by Mario Carneiro, 26-Apr-2015.)
⊢ ◡◡{⟨𝐴, 𝐵⟩} = ◡{⟨𝐵, 𝐴⟩}
Theorem	dmsnsnsn 6173	The domain of the singleton of the singleton of a singleton. (Contributed by NM, 15-Sep-2004.) (Revised by Mario Carneiro, 26-Apr-2015.)
⊢ dom {{{𝐴}}} = {𝐴}
Theorem	rnsnopg 6174	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.)
⊢ (𝐴 ∈ 𝑉 → ran {⟨𝐴, 𝐵⟩} = {𝐵})
Theorem	rnpropg 6175	The range of a pair of ordered pairs is the pair of second members. (Contributed by Thierry Arnoux, 3-Jan-2017.)
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → ran {⟨𝐴, 𝐶⟩, ⟨𝐵, 𝐷⟩} = {𝐶, 𝐷})
Theorem	cnvsng 6176	Converse of a singleton of an ordered pair. (Contributed by NM, 23-Jan-2015.) (Proof shortened by BJ, 12-Feb-2022.)
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → ◡{⟨𝐴, 𝐵⟩} = {⟨𝐵, 𝐴⟩})
Theorem	rnsnop 6177	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}
Theorem	op1sta 6178	Extract the first member of an ordered pair. (See op2nda 6181 to extract the second member, op1stb 5418 for an alternate version, and op1st 7939 for the preferred version.) (Contributed by Raph Levien, 4-Dec-2003.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ ∪ dom {⟨𝐴, 𝐵⟩} = 𝐴
Theorem	cnvsn 6179	Converse of a singleton of an ordered pair. (Contributed by NM, 11-May-1998.) (Revised by Mario Carneiro, 26-Apr-2015.) (Proof shortened by BJ, 12-Feb-2022.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ ◡{⟨𝐴, 𝐵⟩} = {⟨𝐵, 𝐴⟩}
Theorem	op2ndb 6180	Extract the second member of an ordered pair. Theorem 5.12(ii) of [Monk1] p. 52. (See op1stb 5418 to extract the first member, op2nda 6181 for an alternate version, and op2nd 7940 for the preferred version.) (Contributed by NM, 25-Nov-2003.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ ∩ ∩ ∩ ◡{⟨𝐴, 𝐵⟩} = 𝐵
Theorem	op2nda 6181	Extract the second member of an ordered pair. (See op1sta 6178 to extract the first member, op2ndb 6180 for an alternate version, and op2nd 7940 for the preferred version.) (Contributed by NM, 17-Feb-2004.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ ∪ ran {⟨𝐴, 𝐵⟩} = 𝐵
Theorem	opswap 6182	Swap the members of an ordered pair. (Contributed by NM, 14-Dec-2008.) (Revised by Mario Carneiro, 30-Aug-2015.)
⊢ ∪ ◡{⟨𝐴, 𝐵⟩} = ⟨𝐵, 𝐴⟩
Theorem	cnvresima 6183	An image under the converse of a restriction. (Contributed by Jeff Hankins, 12-Jul-2009.)
⊢ (◡(𝐹 ↾ 𝐴) “ 𝐵) = ((◡𝐹 “ 𝐵) ∩ 𝐴)
Theorem	resdm2 6184	A class restricted to its domain equals its double converse. (Contributed by NM, 8-Apr-2007.)
⊢ (𝐴 ↾ dom 𝐴) = ◡◡𝐴
Theorem	resdmres 6185	Restriction to the domain of a restriction. (Contributed by NM, 8-Apr-2007.)
⊢ (𝐴 ↾ dom (𝐴 ↾ 𝐵)) = (𝐴 ↾ 𝐵)
Theorem	resresdm 6186	A restriction by an arbitrary set is a restriction by its domain. (Contributed by AV, 16-Nov-2020.)
⊢ (𝐹 = (𝐸 ↾ 𝐴) → 𝐹 = (𝐸 ↾ dom 𝐹))
Theorem	imadmres 6187	The image of the domain of a restriction. (Contributed by NM, 8-Apr-2007.)
⊢ (𝐴 “ dom (𝐴 ↾ 𝐵)) = (𝐴 “ 𝐵)
Theorem	resdmss 6188	Subset relationship for the domain of a restriction. (Contributed by Scott Fenton, 9-Aug-2024.)
⊢ dom (𝐴 ↾ 𝐵) ⊆ 𝐵
Theorem	resdifdi 6189	Distributive law for restriction over difference. (Contributed by BTernaryTau, 15-Aug-2024.)
⊢ (𝐴 ↾ (𝐵 ∖ 𝐶)) = ((𝐴 ↾ 𝐵) ∖ (𝐴 ↾ 𝐶))
Theorem	resdifdir 6190	Distributive law for restriction over difference. (Contributed by BTernaryTau, 15-Aug-2024.)
⊢ ((𝐴 ∖ 𝐵) ↾ 𝐶) = ((𝐴 ↾ 𝐶) ∖ (𝐵 ↾ 𝐶))
Theorem	mptpreima 6191*	The preimage of a function in maps-to notation. (Contributed by Stefan O'Rear, 25-Jan-2015.)
⊢ 𝐹 = (𝑥 ∈ 𝐴 ↦ 𝐵)    ⇒   ⊢ (◡𝐹 “ 𝐶) = {𝑥 ∈ 𝐴 ∣ 𝐵 ∈ 𝐶}
Theorem	mptiniseg 6192*	Converse singleton image of a function defined by maps-to. (Contributed by Stefan O'Rear, 25-Jan-2015.)
⊢ 𝐹 = (𝑥 ∈ 𝐴 ↦ 𝐵)    ⇒   ⊢ (𝐶 ∈ 𝑉 → (◡𝐹 “ {𝐶}) = {𝑥 ∈ 𝐴 ∣ 𝐵 = 𝐶})
Theorem	dmmpt 6193	The domain of the mapping operation in general. (Contributed by NM, 16-May-1995.) (Revised by Mario Carneiro, 22-Mar-2015.)
⊢ 𝐹 = (𝑥 ∈ 𝐴 ↦ 𝐵)    ⇒   ⊢ dom 𝐹 = {𝑥 ∈ 𝐴 ∣ 𝐵 ∈ V}
Theorem	dmmptss 6194*	The domain of a mapping is a subset of its base class. (Contributed by Scott Fenton, 17-Jun-2013.)
⊢ 𝐹 = (𝑥 ∈ 𝐴 ↦ 𝐵)    ⇒   ⊢ dom 𝐹 ⊆ 𝐴
Theorem	dmmptg 6195*	The domain of the mapping operation is the stated domain, if the function value is always a set. (Contributed by Mario Carneiro, 9-Feb-2013.) (Revised by Mario Carneiro, 14-Sep-2013.)
⊢ (∀𝑥 ∈ 𝐴 𝐵 ∈ 𝑉 → dom (𝑥 ∈ 𝐴 ↦ 𝐵) = 𝐴)
Theorem	rnmpt0f 6196*	The range of a function in maps-to notation is empty if and only if its domain is empty. (Contributed by Glauco Siliprandi, 8-Apr-2021.)
⊢ Ⅎ𝑥𝜑    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝐵 ∈ 𝑉)    &   ⊢ 𝐹 = (𝑥 ∈ 𝐴 ↦ 𝐵)    ⇒   ⊢ (𝜑 → (ran 𝐹 = ∅ ↔ 𝐴 = ∅))
Theorem	rnmptn0 6197*	The range of a function in maps-to notation is nonempty if the domain is nonempty. (Contributed by Glauco Siliprandi, 8-Apr-2021.)
⊢ Ⅎ𝑥𝜑    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝐵 ∈ 𝑉)    &   ⊢ 𝐹 = (𝑥 ∈ 𝐴 ↦ 𝐵)    &   ⊢ (𝜑 → 𝐴 ≠ ∅)    ⇒   ⊢ (𝜑 → ran 𝐹 ≠ ∅)
Theorem	dfco2 6198*	Alternate definition of a class composition, using only one bound variable. (Contributed by NM, 19-Dec-2008.)
⊢ (𝐴 ∘ 𝐵) = ∪ 𝑥 ∈ V ((◡𝐵 “ {𝑥}) × (𝐴 “ {𝑥}))
Theorem	dfco2a 6199*	Generalization of dfco2 6198, where 𝐶 can have any value between dom 𝐴 ∩ ran 𝐵 and V. (Contributed by NM, 21-Dec-2008.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
⊢ ((dom 𝐴 ∩ ran 𝐵) ⊆ 𝐶 → (𝐴 ∘ 𝐵) = ∪ 𝑥 ∈ 𝐶 ((◡𝐵 “ {𝑥}) × (𝐴 “ {𝑥})))
Theorem	coundi 6200	Class composition distributes over union. (Contributed by NM, 21-Dec-2008.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
⊢ (𝐴 ∘ (𝐵 ∪ 𝐶)) = ((𝐴 ∘ 𝐵) ∪ (𝐴 ∘ 𝐶))