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Statement List for Metamath Proof Explorer - 2601-2700 - Page 27 of 58
TypeLabelDescription
Statement
 
Theoremresid 2601 Any relation restricted to the universe is itself.
(Rel A → (AV) = A)
 
Theoremimaeq1 2602 Equality theorem for image.
(A = B → (AC) = (BC))
 
Theoremimaeq2 2603 Equality theorem for image.
(A = B → (CA) = (CB))
 
Theoremdfima2 2604 Alternate definition of image. Compare definition (d) of [Enderton] p. 44.
(AB) = {y∣∃xB xAy}
 
Theoremdfima3 2605 Alternate definition of image. Compare definition (d) of [Enderton] p. 44.
(AB) = {y∣∃x(xB ∧ ⟨x, y⟩ ∈ A)}
 
Theoremelima 2606 Membership in an image. Theorem 34 of [Suppes] p. 65.
AV    ⇒   (A ∈ (BC) ↔ ∃xC xBA)
 
Theoremelima2 2607 Membership in an image. Theorem 34 of [Suppes] p. 65.
AV    ⇒   (A ∈ (BC) ↔ ∃x(xCxBA))
 
Theoremelima3 2608 Membership in an image. Theorem 34 of [Suppes] p. 65.
AV    ⇒   (A ∈ (BC) ↔ ∃x(xC ∧ ⟨x, A⟩ ∈ B))
 
Theoremhbima 2609 Bound-variable hypothesis builder for image.
(yA → ∀x yA)    &   (yB → ∀x yB)    ⇒   (y ∈ (AB) → ∀x y ∈ (AB))
 
Theoremimadmrn 2610 The image of the domain of a class is the range of the class.
(A “ dom A) = ran A
 
Theoremimassrn 2611 The image of a class is a subset of its range. Theorem 3.16(xi) of [Monk1] p. 39.
(AB) ⊆ ran A
 
Theoremimaexg 2612 The image of a set is a set. Theorem 3.17 of [Monk1] p. 39.
(AC → (AB) ∈ V)
 
Theoremimai 2613 Image under the identity relation. Theorem 3.16(viii) of [Monk1] p. 38.
(IA) = A
 
Theoremrnresi 2614 The range of the restricted identity function.
ran (IA) = A
 
Theoremima0 2615 Image of the empty set. Theorem 3.16(ii) of [Monk1] p. 38.
(A “ ∅) = ∅
 
Theoremimasn 2616 Image of a singleton.
(Rel R → (R “ {A}) = {y∣⟨A, y⟩ ∈ R})
 
Theoremelimasn 2617 Membership in an image of a singleton.
BV    &   CV    ⇒   (C ∈ (A “ {B}) ↔ ⟨B, C⟩ ∈ A)
 
Theoremeliniseg 2618 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.
CV    ⇒   (BD → (C ∈ (A “ {B}) ↔ CAB))
 
Theoreminiseg 2619 An idiom that signifies an initial segment of an ordering, used, for example, in Definition 6.21 of [TakeutiZaring] p. 30.
(BC → (A “ {B}) = {xxAB})
 
Theoremdffr3 2620 Alternate definition of founded relation. Definition 6.21 of [TakeutiZaring] p. 30.
(R Fr A ↔ ∀x((xA ∧ ¬ x = ∅) → ∃yx (x ∩ (R “ {y})) = ∅))
 
Theoremimass1 2621 Subset theorem for image.
(AB → (AC) ⊆ (BC))
 
Theoremimass2 2622 Subset theorem for image. Exercise 22(a) of [Enderton] p. 53.
(AB → (CA) ⊆ (CB))
 
Theoremndmima 2623 The image of a singleton outside the domain is empty.
A ∈ dom B → (B “ {A}) = ∅)
 
Theoremrelcnv 2624 A converse is a relation. Theorem 12 of [Suppes] p. 62.
Rel A
 
Theoremcotr 2625 Two ways of saying a relation is transitive. Definition of transitivity in [Schechter] p. 51.
((RR) ⊆ R ↔ ∀xyz((xRyyRz) → xRz))
 
Theoremcnvsym 2626 Two ways of saying a relation is symmetric. Similar to definition of symmetry in [Schechter] p. 51.
(RR ↔ ∀xy(xRyyRx))
 
Theoremintasym 2627 Two ways of saying a relation is antisymmetric. Definition of antisymmetry in [Schechter] p. 51.
((RR) ⊆ I ↔ ∀xy((xRyyRx) → x = y))
 
Theoremintirr 2628 Two ways of saying a relation is irreflexive. Definition of irreflexivity in [Schechter] p. 51.
((RI) = ∅ ↔ ∀x ¬ xRx)
 
Theoremsoirri 2629 A strict order relation is irreflexive.
AV    &   R Or S    &   R ⊆ (S × S)    ⇒    ¬ ARA
 
Theoremsotri 2630 A strict order relation is a transitive relation.
AV    &   R Or S    &   R ⊆ (S × S)    &   BV    &   CV    ⇒   ((ARBBRC) → ARC)
 
Theoremson2lpi 2631 A strict order relation has no 2-cycle loops.
AV    &   R Or S    &   R ⊆ (S × S)    &   BV    ⇒    ¬ (ARBBRA)
 
Theoremcnvopab 2632 The converse of a class abstraction of ordered pairs.
{⟨x, y⟩∣φ} = {⟨y, x⟩∣φ}
 
Theoremcnv0 2633 The converse of the empty set.
∅ = ∅
 
Theoremcnvi 2634 The converse of the identity relation. Theorem 3.7(ii) of [Monk1] p. 36.
I = I
 
Theoremop1sta 2635 Extract the first member of an ordered pair. (See op2nda 2639 to extract the second member and op1stb 1992 for an alternate version.) (Contributed by Raph Levien, 4-Dec-03.)
AV    ⇒   dom {⟨A, B⟩} = A
 
Theoremcnvsn 2636 Converse of a singleton of an ordered pair.
AV    &   BV    ⇒   {⟨A, B⟩} = {⟨B, A⟩}
 
Theoremrnsnop 2637 The range of a singleton of an ordered pair is the singleton of the second member.
AV    &   BV    ⇒   ran {⟨A, B⟩} = {B}
 
Theoremop2ndb 2638 Extract the second member of an ordered pair. Theorem 5.12(ii) of [Monk1] p. 52. (See op1stb 1992 to extract the first member and op2nda 2639 for an alternate version.)
AV    &   BV    ⇒   {⟨A, B⟩} = B
 
Theoremop2nda 2639 Extract the second member of an ordered pair. (See op1sta 2635 to extract the first member and op2ndb 2638 for an alternate version.)
AV    &   BV    ⇒   ran {⟨A, B⟩} = B
 
Theoremelxp4 2640 Membership in a cross product. This version requires no quantifiers or dummy variables. See also elxp5 2641 and elxp6 3093.
(A ∈ (B × C) ↔ (A = ⟨dom {A}, ran {A}⟩ ∧ (dom {A} ∈ Bran {A} ∈ C)))
 
Theoremelxp5 2641 Membership in a cross product requiring no quantifiers or dummy variables. Provides a slightly shorter version of elxp4 2640 when the double intersection does not create class existence problems (caused by int0 1978).
(A ∈ (B × C) ↔ (A = ⟨A, ran {A}⟩ ∧ (ABran {A} ∈ C)))
 
Theoremcnvun 2642 The converse of a union is the union of converses. Theorem 16 of [Suppes] p. 62.
(AB) = (AB)
 
Theoremcnvin 2643 Distributive law for converse over intersection. Theorem 15 of [Suppes] p. 62.
(AB) = (AB)
 
Theoremrnun 2644 Distributive law for range over union. Theorem 8 of [Suppes] p. 60.
ran (AB) = (ran A ∪ ran B)
 
Theoremrnin 2645 The range of an intersection belongs the intersection of ranges. Theorem 9 of [Suppes] p. 60.
ran (AB) ⊆ (ran A ∩ ran B)
 
Theoremrnuni 2646 The range of a union. Part of Exercise 8 of [Enderton] p. 41.
ran A = xA ran x
 
Theoremimaun 2647 Distributive law for image over union. Theorem 35 of [Suppes] p. 65.
(A “ (BC)) = ((AB) ∪ (AC))
 
Theoremdminss 2648 An upper bound for intersection with a domain. Theorem 40 of [Suppes] p. 66, who calls it "somewhat surprising."
(dom RA) ⊆ (R “ (RA))
 
Theoremimainss 2649 An upper bound for intersection with an image. Theorem 41 of [Suppes] p. 66.
((RA) ∩ B) ⊆ (R “ (A ∩ (RB)))
 
Theoremimaiun 2650 The image of a union is the union of the images. Theorem 3K(a) of [Enderton] p. 50.
(AB) = xB (Ax)
 
Theoremcnvxp 2651 The converse of a cross product. Exercise 11 of [Suppes] p. 67.
(A × B) = (B × A)
 
Theoremxp0 2652 The cross product with the empty set is empty. Part of Theorem 3.13(ii) of [Monk1] p. 37.
(A × ∅) = ∅
 
Theoremxpdisj1 2653 Cross products with disjoint sets are disjoint.
((AB) = ∅ → ((A × C) ∩ (B × D)) = ∅)
 
Theoremxpdisj2 2654 Cross products with disjoint sets are disjoint.
((AB) = ∅ → ((C × A) ∩ (D × B)) = ∅)
 
Theoremxpsndisj 2655 Cross products with two different singletons are disjoint.
B = D → ((A × {B}) ∩ (C × {D})) = ∅)
 
Theoremresdisj 2656 A double restriction to disjoint classes is the empty set.
((AB) = ∅ → ((CA) ↾ B) = ∅)
 
Theoremrnxp 2657 The range of a cross product. Part of Theorem 3.13(x) of [Monk1] p. 37.
A = ∅ → ran (A × B) = B)
 
Theoremrelco 2658 A composition is a relation. Exercise 24 of [TakeutiZaring] p. 25.
Rel (AB)
 
Theoremcores 2659 The first member of a composition may be restricted down to the range of the second without affecting the result.
(ran BC → ((AC) ∘ B) = (AB))
 
Theoremdfrel2 2660 Alternate definition of relation. Exercise 2 of [TakeutiZaring] p. 25.
(Rel RR = R)
 
Theoremcnvcnv 2661 The double converse of a class strips out all elements that are not ordered pairs.
A = (A ∩ (V × V))
 
Theoremcnvcnvss 2662 The double converse of a class is a subclass. Exercise 2 of [TakeutiZaring] p. 25.
AA
 
Theoremco02 2663 Composition with the empty set. Theorem 20 of [Suppes] p. 63.
(A ∘ ∅) = ∅
 
Theoremco01 2664 Composition with the empty set.
(∅ ∘ A) = ∅
 
Theoremcoi1 2665 Composition with the identity relation. Part of Theorem 3.7(i) of [Monk1] p. 36.
(Rel A → (AI) = A)
 
Theoremcoi2 2666 Composition with the identity relation. Part of Theorem 3.7(i) of [Monk1] p. 36.
(Rel A → (IA) = A)
 
Theoremcoass 2667 Associative law for class composition. Theorem 27 of [Suppes] p. 64. Also Exercise 21 of [Enderton] p. 53. Interestingly, this law holds for any classes whatsoever, not just functions or even relations.
((AB) ∘ C) = (A ∘ (BC))
 
Theoremrelssdr 2668 A relation is included in the cross product of its domain and range. Exercise 4.12(t) of [Mendelson] p. 235.
(Rel AA ⊆ (dom A × ran A))
 
Theoremcnvexg 2669 The converse of a set is a set. Corollary 6.8(1) of [TakeutiZaring] p. 26.
(ABAV)
 
Theoremcnvex 2670 The converse of a set is a set. Corollary 6.8(1) of [TakeutiZaring] p. 26.
AV    ⇒   AV
 
Theoremcoexg 2671 The composition of two sets is a set.
((ACBD) → (AB) ∈ V)
 
Theoremcoex 2672 The composition of two sets is a set.
AV    &   BV    ⇒   (AB) ∈ V
 
Theoremdmco2 2673 The domain of a composition. Exercise 27 of [Enderton] p. 53.
dom (AB) = (B “ dom A)
 
Theoremdffun2 2674 Alternate definition of a function.
(Fun A ↔ (Rel A ∧ ∀xyz((xAyxAz) → y = z)))
 
Theoremdffun3 2675 Alternate definition of function.
(Fun A ↔ (Rel A ∧ ∀xzy(xAyy = z)))
 
Theoremdffun4 2676 Alternate definition of a function. Definition 6.4(4) of [TakeutiZaring] p. 24.
(Fun A ↔ (Rel A ∧ ∀xyz((⟨x, y⟩ ∈ A ∧ ⟨x, z⟩ ∈ A) → y = z)))
 
Theoremdffun5 2677 Alternate definition of function.
(Fun A ↔ (Rel A ∧ ∀xzy(⟨x, y⟩ ∈ Ay = z)))
 
Theoremdffunmof 2678 Function definition requiring only that x and y not be "free" in A (but not necessarily absent from it), using "at most one" notation.
(zA → ∀x zA)    &   (zA → ∀y zA)    ⇒   (Fun A ↔ (Rel A ∧ ∀x∃*y xAy))
 
Theoremdffunmo 2679 Alternate definition of a function using "at most one" notation.
(Fun A ↔ (Rel A ∧ ∀x∃*y xAy))
 
Theoremfunmo 2680 A function has at most one value for each argument.
(Fun A → ∃*y xAy)
 
Theoremfunrel 2681 A function is a relation.
(Fun A → Rel A)
 
Theoremfunss 2682 Subclass theorem for function predicate.
(AB → (Fun B → Fun A))
 
Theoremfuneq 2683 Equality theorem for function predicate.
(A = B → (Fun A ↔ Fun B))
 
Theoremhbfun 2684 Bound-variable hypothesis builder for a function.
(yF → ∀x yF)    ⇒   (Fun F → ∀xFun F)
 
Theoremfuneu 2685 There is exactly one value of a function.
((Fun FxFy) → ∃!y xFy)
 
Theoremfuneu2 2686 There is exactly one value of a function.
((Fun F ∧ ⟨x, y⟩ ∈ F) → ∃!yx, y⟩ ∈ F)
 
Theoremdffun6 2687 Alternate definition of a function. One possibility for the definition of a function in [Enderton] p. 42. (Enderton's definition is ambiguous because "there is only one" could mean either "there is at most one" or "there is exactly one". However, dffun7 2688 shows that it doesn't matter which meaning we pick.)
(Fun A ↔ (Rel A ∧ ∀x ∈ dom A∃*y xAy))
 
Theoremdffun7 2688 Alternate definition of a function. One possibility for the definition of a function in [Enderton] p. 42. Compare dffun6 2687.
(Fun A ↔ (Rel A ∧ ∀x ∈ dom A∃!y xAy))
 
Theoremfunfn 2689 An equivalence for the function predicate.
(Fun AA Fn dom A)
 
Theoremfunsn 2690 A singleton of an ordered pair is a function.
AV    &   BV    ⇒   Fun {⟨A, B⟩}
 
Theoremfun0 2691 The empty set is a function.
Fun ∅
 
Theoremfuni 2692 The identity relation is a function.
Fun I
 
Theoremnfunv 2693 The universe is not a function. (Contributed by Raph Levien, 27-Jan-04.)
¬ Fun V
 
Theoremfunopab 2694 A class of ordered pairs is a function when there is at most one second member for each pair.
(Fun {⟨x, y⟩∣φ} ↔ ∀x∃*yφ)
 
Theoremfunopabeq 2695 A class of ordered pairs of values is a function.
Fun {⟨x, y⟩∣y = A}
 
Theoremfunco 2696 The composition of two functions is a function. Exercise 29 of [TakeutiZaring] p. 25.
((Fun F ∧ Fun G) → Fun (FG))
 
Theoremfunres 2697 A restriction of a function is a function. Compare Exercise 18 of [TakeutiZaring] p. 25.
(Fun F → Fun (FA))
 
Theoremfunssres 2698 The restriction of a function to the domain of a subclass equals the subclass.
((Fun FGF) → (F ↾ dom G) = G)
 
Theoremfun2ssres 2699 Equality of restrictions of a function and a subclass.
(((Fun FGF) ∧ A ⊆ dom G) → (FA) = (GA))
 
Theoremfunun 2700 The union of functions with disjoint domains is a function. Theorem 4.6 of [Monk1] p. 43.
(((Fun F ∧ Fun G) ∧ (dom F ∩ dom G) = ∅) → Fun (FG))

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