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Theorem List for Intuitionistic Logic Explorer - 4601-4700   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremelima2 4601* Membership in an image. Theorem 34 of [Suppes] p. 65. (Contributed by NM, 11-Aug-2004.)
A V       (A (B𝐶) ↔ x(x 𝐶 xBA))
 
Theoremelima3 4602* Membership in an image. Theorem 34 of [Suppes] p. 65. (Contributed by NM, 14-Aug-1994.)
A V       (A (B𝐶) ↔ x(x 𝐶 x, A B))
 
Theoremnfima 4603 Bound-variable hypothesis builder for image. (Contributed by NM, 30-Dec-1996.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
xA    &   xB       x(AB)
 
Theoremnfimad 4604 Deduction version of bound-variable hypothesis builder nfima 4603. (Contributed by FL, 15-Dec-2006.) (Revised by Mario Carneiro, 15-Oct-2016.)
(φxA)    &   (φxB)       (φx(AB))
 
Theoremimadmrn 4605 The image of the domain of a class is the range of the class. (Contributed by NM, 14-Aug-1994.)
(A “ dom A) = ran A
 
Theoremimassrn 4606 The image of a class is a subset of its range. Theorem 3.16(xi) of [Monk1] p. 39. (Contributed by NM, 31-Mar-1995.)
(AB) ⊆ ran A
 
Theoremimaexg 4607 The image of a set is a set. Theorem 3.17 of [Monk1] p. 39. (Contributed by NM, 24-Jul-1995.)
(A 𝑉 → (AB) V)
 
Theoremimai 4608 Image under the identity relation. Theorem 3.16(viii) of [Monk1] p. 38. (Contributed by NM, 30-Apr-1998.)
( I “ A) = A
 
Theoremrnresi 4609 The range of the restricted identity function. (Contributed by NM, 27-Aug-2004.)
ran ( I ↾ A) = A
 
Theoremresiima 4610 The image of a restriction of the identity function. (Contributed by FL, 31-Dec-2006.)
(BA → (( I ↾ A) “ B) = B)
 
Theoremima0 4611 Image of the empty set. Theorem 3.16(ii) of [Monk1] p. 38. (Contributed by NM, 20-May-1998.)
(A “ ∅) = ∅
 
Theorem0ima 4612 Image under the empty relation. (Contributed by FL, 11-Jan-2007.)
(∅ “ A) = ∅
 
Theoremcsbima12g 4613 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.)
(A 𝐶A / x(𝐹B) = (A / x𝐹A / xB))
 
Theoremimadisj 4614 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 4615 A preimage under any class is included in the domain of the class. (Contributed by FL, 29-Jan-2007.)
(AB) ⊆ dom A
 
Theoremcnvimarndm 4616 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 4617* The image of a singleton. (Contributed by NM, 8-May-2005.)
(A B → (𝑅 “ {A}) = {yA𝑅y})
 
Theoremelreimasng 4618 Elementhood in the image of a singleton. (Contributed by Jim Kingdon, 10-Dec-2018.)
((Rel 𝑅 A 𝑉) → (B (𝑅 “ {A}) ↔ A𝑅B))
 
Theoremelimasn 4619 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 4620 Membership in an image of a singleton. (Contributed by Raph Levien, 21-Oct-2006.)
((B 𝑉 𝐶 𝑊) → (𝐶 (A “ {B}) ↔ ⟨B, 𝐶 A))
 
Theoremargs 4621* 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.)
{xy(𝐹 “ {x}) = {y}} = {x∃!y x𝐹y}
 
Theoremeliniseg 4622 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       (B 𝑉 → (𝐶 (A “ {B}) ↔ 𝐶AB))
 
Theoremepini 4623 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 4624* 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.)
(B 𝑉 → (A “ {B}) = {xxAB})
 
Theoremdfse2 4625* Alternate definition of set-like relation. (Contributed by Mario Carneiro, 23-Jun-2015.)
(𝑅 Se Ax A (A ∩ (𝑅 “ {x})) V)
 
Theoremexse2 4626 Any set relation is set-like. (Contributed by Mario Carneiro, 22-Jun-2015.)
(𝑅 𝑉𝑅 Se A)
 
Theoremimass1 4627 Subset theorem for image. (Contributed by NM, 16-Mar-2004.)
(AB → (A𝐶) ⊆ (B𝐶))
 
Theoremimass2 4628 Subset theorem for image. Exercise 22(a) of [Enderton] p. 53. (Contributed by NM, 22-Mar-1998.)
(AB → (𝐶A) ⊆ (𝐶B))
 
Theoremndmima 4629 The image of a singleton outside the domain is empty. (Contributed by NM, 22-May-1998.)
A dom B → (B “ {A}) = ∅)
 
Theoremrelcnv 4630 A converse is a relation. Theorem 12 of [Suppes] p. 62. (Contributed by NM, 29-Oct-1996.)
Rel A
 
Theoremrelbrcnvg 4631 When 𝑅 is a relation, the sethood assumptions on brcnv 4445 can be omitted. (Contributed by Mario Carneiro, 28-Apr-2015.)
(Rel 𝑅 → (A𝑅BB𝑅A))
 
Theoremrelbrcnv 4632 When 𝑅 is a relation, the sethood assumptions on brcnv 4445 can be omitted. (Contributed by Mario Carneiro, 28-Apr-2015.)
Rel 𝑅       (A𝑅BB𝑅A)
 
Theoremcotr 4633* 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) ⊆ 𝑅xyz((x𝑅y y𝑅z) → x𝑅z))
 
Theoremissref 4634* 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) ⊆ 𝑅x A x𝑅x)
 
Theoremcnvsym 4635* 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.)
(𝑅𝑅xy(x𝑅yy𝑅x))
 
Theoremintasym 4636* 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((x𝑅y y𝑅x) → x = y))
 
Theoremasymref 4637* Two ways of saying a relation is antisymmetric and reflexive. 𝑅 is the field of a relation by relfld 4773. (Contributed by NM, 6-May-2008.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)
((𝑅𝑅) = ( I ↾ 𝑅) ↔ x 𝑅y((x𝑅y y𝑅x) ↔ x = y))
 
Theoremintirr 4638* 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 4639* Two ways of saying that two elements have an upper bound. (Contributed by Mario Carneiro, 3-Nov-2015.)
((A 𝑉 B 𝑊) → (A(𝑅𝑅)Bz(A𝑅z B𝑅z)))
 
Theoremcodir 4640* Two ways of saying a relation is directed. (Contributed by Mario Carneiro, 22-Nov-2013.)
((A × B) ⊆ (𝑅𝑅) ↔ x A y B z(x𝑅z y𝑅z))
 
Theoremqfto 4641* A quantifier-free way of expressing the total order predicate. (Contributed by Mario Carneiro, 22-Nov-2013.)
((A × B) ⊆ (𝑅𝑅) ↔ x A y B (x𝑅y y𝑅x))
 
Theoremxpidtr 4642 A square cross product (A × A) is a transitive relation. (Contributed by FL, 31-Jul-2009.)
((A × A) ∘ (A × A)) ⊆ (A × A)
 
Theoremtrin2 4643 The intersection of two transitive classes is transitive. (Contributed by FL, 31-Jul-2009.)
(((𝑅𝑅) ⊆ 𝑅 (𝑆𝑆) ⊆ 𝑆) → ((𝑅𝑆) ∘ (𝑅𝑆)) ⊆ (𝑅𝑆))
 
Theorempoirr2 4644 A partial order relation is irreflexive. (Contributed by Mario Carneiro, 2-Nov-2015.)
(𝑅 Po A → (𝑅 ∩ ( I ↾ A)) = ∅)
 
Theoremtrinxp 4645 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 4646 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 4647 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)       ((A𝑅B B𝑅𝐶) → A𝑅𝐶)
 
Theoremson2lpi 4648 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)        ¬ (A𝑅B B𝑅A)
 
Theoremsotri2 4649 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)       ((A 𝑆 ¬ B𝑅A B𝑅𝐶) → A𝑅𝐶)
 
Theoremsotri3 4650 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)       ((𝐶 𝑆 A𝑅B ¬ 𝐶𝑅B) → A𝑅𝐶)
 
Theorempoleloe 4651 Express "less than or equals" for general strict orders. (Contributed by Stefan O'Rear, 17-Jan-2015.)
(B 𝑉 → (A(𝑅 ∪ I )B ↔ (A𝑅B A = B)))
 
Theorempoltletr 4652 Transitive law for general strict orders. (Contributed by Stefan O'Rear, 17-Jan-2015.)
((𝑅 Po 𝑋 (A 𝑋 B 𝑋 𝐶 𝑋)) → ((A𝑅B B(𝑅 ∪ I )𝐶) → A𝑅𝐶))
 
Theoremcnvopab 4653* 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}
 
Theoremcnv0 4654 The converse of the empty set. (Contributed by NM, 6-Apr-1998.)
∅ = ∅
 
Theoremcnvi 4655 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 4656 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 4657 Distributive law for converse over set difference. (Contributed by Mario Carneiro, 26-Jun-2014.)
(AB) = (AB)
 
Theoremcnvin 4658 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 4659 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 4660 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 4661 The range of an indexed union. (Contributed by Mario Carneiro, 29-May-2015.)
ran x A B = x A ran B
 
Theoremrnuni 4662* 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
 
Theoremimaundi 4663 Distributive law for image over union. Theorem 35 of [Suppes] p. 65. (Contributed by NM, 30-Sep-2002.)
(A “ (B𝐶)) = ((AB) ∪ (A𝐶))
 
Theoremimaundir 4664 The image of a union. (Contributed by Jeff Hoffman, 17-Feb-2008.)
((AB) “ 𝐶) = ((A𝐶) ∪ (B𝐶))
 
Theoremdminss 4665 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 4666 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 4667 The image of an intersection (Contributed by Thierry Arnoux, 16-Dec-2017.)
((AB) “ 𝐶) ⊆ ((A𝐶) ∩ (B𝐶))
 
Theoreminimasn 4668 The intersection of the image of singleton (Contributed by Thierry Arnoux, 16-Dec-2017.)
(𝐶 𝑉 → ((AB) “ {𝐶}) = ((A “ {𝐶}) ∩ (B “ {𝐶})))
 
Theoremcnvxp 4669 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 4670 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 4671* The cross product of inhabited classes is inhabited. (Contributed by Jim Kingdon, 11-Dec-2018.)
((x x A y y B) ↔ z z (A × B))
 
Theoremxpm 4672* The cross product of inhabited classes is inhabited. (Contributed by Jim Kingdon, 13-Dec-2018.)
((x x A y y B) ↔ z z (A × B))
 
Theoremxpeq0r 4673 A cross product is empty if at least one member is empty. (Contributed by Jim Kingdon, 12-Dec-2018.)
((A = ∅ B = ∅) → (A × B) = ∅)
 
Theoremxpdisj1 4674 Cross products with disjoint sets are disjoint. (Contributed by NM, 13-Sep-2004.)
((AB) = ∅ → ((A × 𝐶) ∩ (B × 𝐷)) = ∅)
 
Theoremxpdisj2 4675 Cross products with disjoint sets are disjoint. (Contributed by NM, 13-Sep-2004.)
((AB) = ∅ → ((𝐶 × A) ∩ (𝐷 × B)) = ∅)
 
Theoremxpsndisj 4676 Cross products with two different singletons are disjoint. (Contributed by NM, 28-Jul-2004.)
(B𝐷 → ((A × {B}) ∩ (𝐶 × {𝐷})) = ∅)
 
Theoremdjudisj 4677* Disjoint unions with disjoint index sets are disjoint. (Contributed by Stefan O'Rear, 21-Nov-2014.)
((AB) = ∅ → ( x A ({x} × 𝐶) ∩ y B ({y} × 𝐷)) = ∅)
 
Theoremresdisj 4678 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 4679* The range of a cross product. Part of Theorem 3.13(x) of [Monk1] p. 37, with non-empty changed to inhabited. (Contributed by Jim Kingdon, 12-Dec-2018.)
(x x A → ran (A × B) = B)
 
Theoremdmxpss 4680 The domain of a cross product is a subclass of the first factor. (Contributed by NM, 19-Mar-2007.)
dom (A × B) ⊆ A
 
Theoremrnxpss 4681 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
 
Theoremrnxpid 4682 The range of a square cross product. (Contributed by FL, 17-May-2010.)
ran (A × A) = A
 
Theoremssxpbm 4683* 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) ↔ (A𝐶 B𝐷)))
 
Theoremssxp1 4684* Cross product subset cancellation. (Contributed by Jim Kingdon, 14-Dec-2018.)
(x x 𝐶 → ((A × 𝐶) ⊆ (B × 𝐶) ↔ AB))
 
Theoremssxp2 4685* Cross product subset cancellation. (Contributed by Jim Kingdon, 14-Dec-2018.)
(x x 𝐶 → ((𝐶 × A) ⊆ (𝐶 × B) ↔ AB))
 
Theoremxp11m 4686* The cross product of inhabited classes is one-to-one. (Contributed by Jim Kingdon, 13-Dec-2018.)
((x x A y y B) → ((A × B) = (𝐶 × 𝐷) ↔ (A = 𝐶 B = 𝐷)))
 
Theoremxpcanm 4687* Cancellation law for cross-product. (Contributed by Jim Kingdon, 14-Dec-2018.)
(x x 𝐶 → ((𝐶 × A) = (𝐶 × B) ↔ A = B))
 
Theoremxpcan2m 4688* Cancellation law for cross-product. (Contributed by Jim Kingdon, 14-Dec-2018.)
(x x 𝐶 → ((A × 𝐶) = (B × 𝐶) ↔ A = B))
 
Theoremxpexr2m 4689* 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 4690 Subset of the range of a restriction. (Contributed by NM, 16-Jan-2006.)
(B ⊆ ran (𝐶A) ↔ ran (𝐶 ∩ (A × B)) = B)
 
Theoremrninxp 4691* 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)) = By B x A x𝐶y)
 
Theoremdminxp 4692* Domain of the intersection with a cross product. (Contributed by NM, 17-Jan-2006.)
(dom (𝐶 ∩ (A × B)) = Ax A y B x𝐶y)
 
Theoremimainrect 4693 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 4694 The image by a cross product. (Contributed by Thierry Arnoux, 16-Dec-2017.)
((A𝐶) = ∅ → ((A × B) “ 𝐶) = ∅)
 
Theoremxpima2m 4695* The image by a cross product. (Contributed by Thierry Arnoux, 16-Dec-2017.)
(x x (A𝐶) → ((A × B) “ 𝐶) = B)
 
Theoremxpimasn 4696 The image of a singleton by a cross product. (Contributed by Thierry Arnoux, 14-Jan-2018.)
(𝑋 A → ((A × B) “ {𝑋}) = B)
 
Theoremcnvcnv3 4697* 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 4698 Alternate definition of relation. Exercise 2 of [TakeutiZaring] p. 25. (Contributed by NM, 29-Dec-1996.)
(Rel 𝑅𝑅 = 𝑅)
 
Theoremdfrel4v 4699* 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 4700 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))
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