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Statement List for Metamath Proof Explorer - 3201-3300 - Page 33 of 58
TypeLabelDescription
Statement
 
Theoremer2 3201 Alternate definition of equivalence predicate.
(Er R ↔ ∀xyz((xRyyRx) ∧ ((xRyyRz) → xRz)))
 
Definitiondf-ec 3202 Define the R -coset of A. Exercise 35 of [Enderton] p. 61. This is called the equivalence class of A modulo R when R is an equivalence relation. The Definition of [Enderton] p. 57 uses the notation [A] (subscript) R, although we simply follow the brackets by R since we don't have subscripts. For an alternate definition, see ec2 3203.
[A]R = (R “ {A})
 
Theoremec2 3203 Alternate definition of R-coset of A. Definition 34 of [Suppes] p. 81.
AV    ⇒   [A]R = {yARy}
 
Theoremecexg 3204 An equivalence class modulo a set is a set.
(RB → [A]RV)
 
Definitiondf-qs 3205 Define quotient set. R is usually an equivalence relation. Definition of [Enderton] p. 58.
(A / R) = {y∣∃xA y = [x]R}
 
Theoremereq 3206 Equality theorem for equivalence predicate.
(R = S → (Er R ↔ Er S))
 
Theoremster 3207 A symmetric, transitive relation is an equivalence relation.
(xRyyRx)    &   ((xRyyRz) → xRz)    ⇒   Er R
 
Theoremider 3208 The identity relation is an equivalence relation.
Er I
 
Theoremersym 3209 An equivalence relation is symmetric.
AV    &   BV    &   Er R    ⇒   (ARBBRA)
 
Theoremersymb 3210 An equivalence relation is symmetric.
AV    &   BV    &   Er R    ⇒   (ARBBRA)
 
Theoremertr 3211 An equivalence relation is transitive.
AV    &   BV    &   CV    &   Er R    ⇒   ((ARBBRC) → ARC)
 
Theoremerref 3212 An equivalence relation is reflexive on its field. Compare Theorem 3M of [Enderton] p. 56.
Er R    ⇒   (A ∈ (dom R ∪ ran R) → ARA)
 
Theoremerdmrn 3213 The range and domain of an equivalence relation are equal.
Er R    ⇒   dom R = ran R
 
Theoremeceq1 3214 Equality theorem for equivalence class.
(A = B → [C]A = [C]B)
 
Theoremeceq2 3215 Equality theorem for equivalence class.
(A = B → [A]C = [B]C)
 
Theoremelec 3216 Membership in an equivalence class. Theorem 72 of [Suppes] p. 82.
AV    &   BV    ⇒   (A ∈ [B]RBRA)
 
Theoremecdmn0 3217 An equivalence class is not empty in its domain.
AV    ⇒   (A ∈ dom R ↔ ¬ [A]R = ∅)
 
Theoremerthi 3218 Basic property of equivalence relations. Part of Lemma 3N of [Enderton] p. 57.
AV    &   BV    &   Er R    ⇒   (ARB → [A]R = [B]R)
 
Theoremerth 3219 Basic property of equivalence relations. Theorem 73 of [Suppes] p. 82.
BV    &   Er R    ⇒   (A ∈ (dom R ∪ ran R) → ([A]R = [B]RARB))
 
Theoremerthdm 3220 Basic property of equivalence relations. Compare Theorem 73 of [Suppes] p. 82. Assumes membership in the domain instead of just the field.
BV    &   Er R    ⇒   (A ∈ dom R → ([A]R = [B]RARB))
 
Theoremerthdmr 3221 Basic property of equivalence relations. Compare Theorem 73 of [Suppes] p. 82. Assumes membership of the second argument in the domain.
AV    &   BV    &   Er R    ⇒   (B ∈ dom R → ([A]R = [B]RARB))
 
Theoremereldm 3222 Equality of equivalence classes implies equivalence of domain membership.
AV    &   BV    &   Er R    &   dom R = D    ⇒   ([A]R = [B]R → (ADBD))
 
Theoremerdisj 3223 Equivalence classes do not overlap. In other words, two equivalence classes are either equal or disjoint. Theorem 74 of [Suppes] p. 83.
AV    &   BV    &   Er R    ⇒   ([A]R = [B]R ∨ ([A]R ∩ [B]R) = ∅)
 
Theoremecidsn 3224 An equivalence class modulo the identity relation is a singleton.
[A]I = {A}
 
Theoremqseq1 3225 Equality theorem for quotient set.
(A = B → (A / C) = (B / C))
 
Theoremqseq2 3226 Equality theorem for quotient set.
(A = B → (C / A) = (C / B))
 
Theoremelqs 3227 Membership in a quotient set.
BV    ⇒   (B ∈ (A / R) ↔ ∃x(xAB = [x]R))
 
Theoremelqsi 3228 Membership in a quotient set.
(B ∈ (A / R) → ∃x(xAB = [x]R))
 
Theoremecelqsi 3229 Membership of an equivalence class in a quotient set.
RV    ⇒   (BA → [B]R ∈ (A / R))
 
Theoremecopqsi 3230 "Closure" law for equivalence class of ordered pairs.
RV    &   S = ((A × A) / R)    ⇒   ((BACA) → [⟨B, C⟩]RS)
 
Theoremqsex 3231 A quotient set exists.
AV    ⇒   (A / R) ∈ V
 
Theoremsnec 3232 The singleton of an equivalence class.
AV    ⇒   {[A]R} = ({A} / R)
 
Theoremecqs 3233 Equivalence class in terms of quotient set.
AV    &   RV    ⇒   [A]R = ({A} / R)
 
Theorem0nelqs 3234 A quotient set doesn't contain the empty set.
Er R    &   dom R = A    ⇒    ¬ ∅ ∈ (A / R)
 
Theoremecelqsdm 3235 Membership of an equivalence class in a quotient set.
BV    &   Er R    &   dom R = A    ⇒   ([B]R ∈ (A / R) → BA)
 
Theoremecid 3236 A set is equal to its converse epsilon coset. (Note: converse epsilon is not an equivalence relation.)
AV    ⇒   [A]E = A
 
Theoremqsid 3237 A set is equal to its quotient set mod converse epsilon. (Note: converse epsilon is not an equivalence relation.)
AV    ⇒   (A / E) = A
 
Theoremectocl 3238 Implicit substitution of class for equivalence class.
S = (B / R)    &   ([x]R = A → (φψ))    &   (xBφ)    ⇒   (ASψ)
 
Theoremecoptocl 3239 Implicit substitution of class for equivalence class of ordered pair.
S = ((B × C) / R)    &   ([⟨x, y⟩]R = A → (φψ))    &   ((xByC) → φ)    ⇒   (ASψ)
 
Theorem2ecoptocl 3240 Implicit substitution of classes for equivalence classes of ordered pairs.
S = ((C × D) / R)    &   ([⟨x, y⟩]R = A → (φψ))    &   ([⟨z, w⟩]R = B → (ψχ))    &   (((xCyD) ∧ (zCwD)) → φ)    ⇒   ((ASBS) → χ)
 
Theorem3ecoptocl 3241 Implicit substitution of classes for equivalence classes of ordered pairs.
S = ((D × D) / R)    &   ([⟨x, y⟩]R = A → (φψ))    &   ([⟨z, w⟩]R = B → (ψχ))    &   ([⟨v, u⟩]R = C → (χθ))    &   (((xDyD) ∧ (zDwD) ∧ (vDuD)) → φ)    ⇒   ((ASBSCS) → θ)
 
Theorembrecop 3242 Binary relation on a quotient set. Lemma for real number construction.
SV    &   Er S    &   dom S = (G × G)    &   H = ((G × G) / S)    &   R = {⟨x, y⟩∣((xHyH) ∧ ∃zwvu((x = [⟨z, w⟩]Sy = [⟨v, u⟩]S) ∧ φ))}    &   ((((zGwG) ∧ (AGBG)) ∧ ((vGuG) ∧ (CGDG))) → (([⟨z, w⟩]S = [⟨A, B⟩]S ∧ [⟨v, u⟩]S = [⟨C, D⟩]S) → (φψ)))    ⇒   (((AGBG) ∧ (CGDG)) → ([⟨A, B⟩]SR[⟨C, D⟩]Sψ))
 
Theorembrecop2 3243 Binary relation on a quotient set. Lemma for real number construction. Eliminates antecedent from last hypothesis.
SV    &   BV    &   CV    &   DV    &   Er S    &   dom S = (G × G)    &   H = ((G × G) / S)    &   R ⊆ (H × H)    &   Q ⊆ (G × G)    &    ¬ ∅ ∈ G    &   dom F = (G × G)    &   (((AGBG) ∧ (CGDG)) → ([⟨A, B⟩]SR[⟨C, D⟩]S ↔ (AFD)Q(BFC)))    ⇒   ([⟨A, B⟩]SR[⟨C, D⟩]S ↔ (AFD)Q(BFC))
 
Theoremecopopreq 3244 Express the relation R (specified by the hypothesis) in terms of its operation F.
R = {⟨x, y⟩∣((x ∈ (S × S) ∧ y ∈ (S × S)) ∧ ∃zwvu((x = ⟨z, w⟩ ∧ y = ⟨v, u⟩) ∧ (zFu) = (wFv)))}    ⇒   (((ASBS) ∧ (CSDS)) → (⟨A, BRC, D⟩ ↔ (AFD) = (BFC)))
 
Theoremecopoprdm 3245 Assuming the operation F is commutative, compute the domain the relation R specified by the first hypothesis.
R = {⟨x, y⟩∣((x ∈ (S × S) ∧ y ∈ (S × S)) ∧ ∃zwvu((x = ⟨z, w⟩ ∧ y = ⟨v, u⟩) ∧ (zFu) = (wFv)))}    &   (xFy) = (yFx)    ⇒   dom R = (S × S)
 
Theoremecopoprsym 3246 Assuming the operation F is commutative, show that the relation R, specified by the first hypothesis, is symmetric.
R = {⟨x, y⟩∣((x ∈ (S × S) ∧ y ∈ (S × S)) ∧ ∃zwvu((x = ⟨z, w⟩ ∧ y = ⟨v, u⟩) ∧ (zFu) = (wFv)))}    &   (xFy) = (yFx)    &   BV    ⇒   (ARBBRA)
 
Theoremecopoprtrn 3247 Assuming that operation F is commutative (second hypothesis), closed (third hypothesis), associative (fourth hypothesis), and has the cancellation property (fifth hypothesis), show that the relation R, specified by the first hypothesis, is transitive.
R = {⟨x, y⟩∣((x ∈ (S × S) ∧ y ∈ (S × S)) ∧ ∃zwvu((x = ⟨z, w⟩ ∧ y = ⟨v, u⟩) ∧ (zFu) = (wFv)))}    &   (xFy) = (yFx)    &   ((xSyS) → (xFy) ∈ S)    &   ((xFy)Fz) = (xF(yFz))    &   ((xSyS) → ((xFy) = (xFz) → y = z))    &   BV    &   CV    ⇒   ((ARBBRC) → ARC)
 
Theoremecopoprer 3248 Assuming that operation F is commutative (second hypothesis), closed (third hypothesis), associative (fourth hypothesis), and has the cancellation property (fifth hypothesis), show that the relation R, specified by the first hypothesis, is an equivalence relation.
R = {⟨x, y⟩∣((x ∈ (S × S) ∧ y ∈ (S × S)) ∧ ∃zwvu((x = ⟨z, w⟩ ∧ y = ⟨v, u⟩) ∧ (zFu) = (wFv)))}    &   (xFy) = (yFx)    &   ((xSyS) → (xFy) ∈ S)    &   ((xFy)Fz) = (xF(yFz))    &   ((xSyS) → ((xFy) = (xFz) → y = z))    ⇒   Er R
 
Theoremeceqopreq 3249 Equality of equivalence relation in terms of an operation.
BV    &   CV    &   DV    &   Er R    &   dom R = (S × S)    &   dom F = (S × S)    &    ¬ ∅ ∈ S    &   ((xSyS) → (xFy) ∈ S)    &   (((ASBS) ∧ (CSDS)) → (⟨A, BRC, D⟩ ↔ (AFD) = (BFC)))    ⇒   ((ASCS) → ([⟨A, B⟩]R = [⟨C, D⟩]R ↔ (AFD) = (BFC)))
 
Theoremth3qlem1 3250 Lemma for Exercise 44 version of Theorem 3Q of [Enderton] p. 60. The third hypothesis is the compatibility assumption.
Er R    &   dom R = S    &   (((ySwS) ∧ (zSvS)) → ((yRwzRv) → (yFz)R(wFv)))    ⇒   ((A ∈ (S / R) ∧ B ∈ (S / R)) → ∃*xyz((A = [y]RB = [z]R) ∧ x = [(yFz)]R))
 
Theoremth3qlem2 3251 Lemma for Exercise 44 version of Theorem 3Q of [Enderton] p. 60, extended to operations on ordered pairs. The fourth hypothesis is the compatibility assumption.
RV    &   Er R    &   dom R = (S × S)    &   ((((wSvS) ∧ (uStS)) ∧ ((sSfS) ∧ (gShS))) → ((⟨w, vRu, t⟩ ∧ ⟨s, fRg, h⟩) → (⟨w, vFs, f⟩)R(⟨u, tFg, h⟩)))    ⇒   ((A ∈ ((S × S) / R) ∧ B ∈ ((S × S) / R)) → ∃*zwvut((A = [⟨w, v⟩]RB = [⟨u, t⟩]R) ∧ z = [(⟨w, vFu, t⟩)]R))
 
Theoremth3qcor 3252 Corollary of Theorem 3Q of [Enderton] p. 60.
RV    &   Er R    &   dom R = (S × S)    &   ((((wSvS) ∧ (uStS)) ∧ ((sSfS) ∧ (gShS))) → ((⟨w, vRu, t⟩ ∧ ⟨s, fRg, h⟩) → (⟨w, vFs, f⟩)R(⟨u, tFg, h⟩)))    &   G = {⟨⟨x, y⟩, z⟩∣((x ∈ ((S × S) / R) ∧ y ∈ ((S × S) / R)) ∧ ∃wvut((x = [⟨w, v⟩]Ry = [⟨u, t⟩]R) ∧ z = [(⟨w, vFu, t⟩)]R))}    ⇒   Fun G
 
Theoremth3q 3253 Theorem 3Q of [Enderton] p. 60, extended to operations on ordered pairs.
RV    &   Er R    &   dom R = (S × S)    &   ((((wSvS) ∧ (uStS)) ∧ ((sSfS) ∧ (gShS))) → ((⟨w, vRu, t⟩ ∧ ⟨s, fRg, h⟩) → (⟨w, vFs, f⟩)R(⟨u, tFg, h⟩)))    &   G = {⟨⟨x, y⟩, z⟩∣((x ∈ ((S × S) / R) ∧ y ∈ ((S × S) / R)) ∧ ∃wvut((x = [⟨w, v⟩]Ry = [⟨u, t⟩]R) ∧ z = [(⟨w, vFu, t⟩)]R))}    ⇒   (((ASBS) ∧ (CSDS)) → ([⟨A, B⟩]RG[⟨C, D⟩]R) = [(⟨A, BFC, D⟩)]R)
 
Theoremoprec 3254 Express an operation on equivalence classes of ordered pairs in terms of equivalence class of operations on ordered pairs.
HV    &   KV    &   LV    &   RV    &   Er R    &   dom R = (S × S)    &   R = {⟨x, y⟩∣((x ∈ (S × S) ∧ y ∈ (S × S)) ∧ ∃zwvu((x = ⟨z, w⟩ ∧ y = ⟨v, u⟩) ∧ φ))}    &   (((z = aw = b) ∧ (v = cu = d)) → (φψ))    &   (((z = gw = h) ∧ (v = tu = s)) → (φχ))    &   G = {⟨⟨x, y⟩, z⟩∣((x ∈ (S × S) ∧ y ∈ (S × S)) ∧ ∃wvuf((x = ⟨w, v⟩ ∧ y = ⟨u, f⟩) ∧ z = J))}    &   (((w = av = b) ∧ (u = gf = h)) → J = K)    &   (((w = cv = d) ∧ (u = tf = s)) → J = L)    &   (((w = Av = B) ∧ (u = Cf = D)) → J = H)    &   F = {⟨⟨x, y⟩, z⟩∣((xQyQ) ∧ ∃abcd((x = [⟨a, b⟩]Ry = [⟨c, d⟩]R) ∧ z = [(⟨a, bGc, d⟩)]R))}    &   Q = ((S × S) / R)    &   ((((aSbS) ∧ (cSdS)) ∧ ((gShS) ∧ (tSsS))) → ((ψχ) → KRL))    ⇒   (((ASBS) ∧ (CSDS)) → ([⟨A, B⟩]RF[⟨C, D⟩]R) = [H]R)
 
Theoremecoprcom 3255 Lemma used in proving commutative laws via equivalence classes.
C = ((S × S) / R)    &   (((xSyS) ∧ (zSwS)) → ([⟨x, y⟩]RF[⟨z, w⟩]R) = [⟨D, G⟩]R)    &   (((zSwS) ∧ (xSyS)) → ([⟨z, w⟩]RF[⟨x, y⟩]R) = [⟨H, J⟩]R)    &   D = H    &   G = J    ⇒   ((ACBC) → (AFB) = (BFA))
 
Theoremecoprass 3256 Lemma used in proving associative laws via equivalence classes.
D = ((S × S) / R)    &   (((xSyS) ∧ (zSwS)) → ([⟨x, y⟩]RF[⟨z, w⟩]R) = [⟨G, H⟩]R)    &   (((zSwS) ∧ (vSuS)) → ([⟨z, w⟩]RF[⟨v, u⟩]R) = [⟨N, Q⟩]R)    &   (((GSHS) ∧ (vSuS)) → ([⟨G, H⟩]RF[⟨v, u⟩]R) = [⟨J, K⟩]R)    &   (((xSyS) ∧ (NSQS)) → ([⟨x, y⟩]RF[⟨N, Q⟩]R) = [⟨L, M⟩]R)    &   (((xSyS) ∧ (zSwS)) → (GSHS))    &   (((zSwS) ∧ (vSuS)) → (NSQS))    &   J = L    &   K = M    ⇒   ((ADBDCD) → ((AFB)FC) = (AF(BFC)))
 
Theoremecoprdi 3257 Lemma used in proving distributive laws via equivalence classes.
D = ((S × S) / R)    &   (((zSwS) ∧ (vSuS)) → ([⟨z, w⟩]RF[⟨v, u⟩]R) = [⟨M, N⟩]R)    &   (((xSyS) ∧ (MSNS)) → ([⟨x, y⟩]RG[⟨M, N⟩]R) = [⟨H, J⟩]R)    &   (((xSyS) ∧ (zSwS)) → ([⟨x, y⟩]RG[⟨z, w⟩]R) = [⟨W, X⟩]R)    &   (((xSyS) ∧ (vSuS)) → ([⟨x, y⟩]RG[⟨v, u⟩]R) = [⟨Y, Z⟩]R)    &   (((WSXS) ∧ (YSZS)) → ([⟨W, X⟩]RF[⟨Y, Z⟩]R) = [⟨K, L⟩]R)    &   (((zSwS) ∧ (vSuS)) → (MSNS))    &   (((xSyS) ∧ (zSwS)) → (WSXS))    &   (((xSyS) ∧ (vSuS)) → (YSZS))    &   H = K    &   J = L    ⇒   ((ADBDCD) → (AG(BFC)) = ((AGB)F(AGC)))
 
Syntaxcm 3258 Extend the definition of a class to include the mapping operation. (Read for Am B, "the set of all functions that map from B to A.)
class m
 
Definitiondf-map 3259 Define the mapping operation or set exponentiation. The set of all functions that map from B to A is written (Am B) (see mapval 3264). Many authors write A followed by B as a superscript for this operation and rely on context to avoid confusion other exponentiation operations (e.g. Definition 10.42 of [TakeutiZaring] p. 95). Other authors show B as a prefixed superscript, which is read "A pre B" (e.g. definition of [Enderton] p. 52). Definition 8.21 of [Eisenberg] p. 125 uses the notation Map(B, A) for our (Am B). The up-arrow is used by Donald Knuth for iterated exponentiation (Science 194, 1235-1242, 1976). We adopt the first case of his notation (simple exponentiation) and subscript it with m to distinguish it from other kinds of exponentiation.
m = {⟨⟨x, y⟩, z⟩∣z = {ff:y–→x}}
 
Theoremmapprc 3260 When A is a proper class, the class of all functions mapping A to B is empty. Exercise 4.41 of [Mendelson] p. 255.
AV → {ff:A–→B} = ∅)
 
Theoremmapex 3261 The class of all functions mapping one set to another is a set. Remark after Definition 10.24 of [Kunen] p. 31. (Contributed by Raph Levien, 4-Dec-03.)
((ACBD) → {ff:A–→B} ∈ V)
 
Theoremfnmap 3262 Set exponentiation has a universal domain.
m Fn (V × V)
 
Theoremmapvalg 3263 The value of set exponentiation. (Am B) is the set of all functions that map from B to A. Definition 10.24 of [Kunen] p. 24.
((ACBD) → (Am B) = {ff:B–→A})
 
Theoremmapval 3264 The value of set exponentiation (inference version). (Am B) is the set of all functions that map from B to A. Definition 10.24 of [Kunen] p. 24.
AV    &   BV    ⇒   (Am B) = {ff:B–→A}
 
Theoremelmap 3265 Membership relation for set exponentiation.
AV    &   BV    ⇒   (C ∈ (Am B) ↔ C:B–→A)
 
Theoremmap0e 3266 Set exponentiation with an empty exponent (ordinal number 0) is ordinal number 1. Exercise 4.42(a) of [Mendelson] p. 255.
AV    ⇒   (Am ∅) = 1o
 
Theoremmap0b 3267 Set exponentiation with an empty base is the empty set, provided the exponent is non-empty. Theorem 96 of [Suppes] p. 89.
AV    ⇒   A = ∅ → (∅ ↑m A) = ∅)
 
Theoremmap0 3268 Set exponentiation is empty iff the base is empty and the exponent is not empty. Theorem 97 of [Suppes] p. 89.
AV    &   BV