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Theorem List for Intuitionistic Logic Explorer - 6801-6900   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremnnaordex 6801* Equivalence for ordering. Compare Exercise 23 of [Enderton] p. 88. (Contributed by NM, 5-Dec-1995.) (Revised by Mario Carneiro, 15-Nov-2014.)
((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (𝐴 ∈ 𝐵 ↔ ∃𝑥 ∈ ω (∅ ∈ 𝑥 ∧ (𝐴 +o 𝑥) = 𝐵)))
 
Theoremnnawordex 6802* Equivalence for weak ordering of natural numbers. (Contributed by NM, 8-Nov-2002.) (Revised by Mario Carneiro, 15-Nov-2014.)
((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (𝐴 ⊆ 𝐵 ↔ ∃𝑥 ∈ ω (𝐴 +o 𝑥) = 𝐵))
 
Theoremnnm00 6803 The product of two natural numbers is zero iff at least one of them is zero. (Contributed by Jim Kingdon, 11-Nov-2004.)
((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → ((𝐴 ·o 𝐵) = ∅ ↔ (𝐴 = ∅ ∨ 𝐵 = ∅)))
 
2.6.26  Equivalence relations and classes
 
Syntaxwer 6804 Extend the definition of a wff to include the equivalence predicate.
wff 𝑅 Er 𝐴
 
Syntaxcec 6805 Extend the definition of a class to include equivalence class.
class [𝐴]𝑅
 
Syntaxcqs 6806 Extend the definition of a class to include quotient set.
class (𝐴 / 𝑅)
 
Definitiondf-er 6807 Define the equivalence relation predicate. Our notation is not standard. A formal notation doesn't seem to exist in the literature; instead only informal English tends to be used. The present definition, although somewhat cryptic, nicely avoids dummy variables. In dfer2 6808 we derive a more typical definition. We show that an equivalence relation is reflexive, symmetric, and transitive in erref 6827, ersymb 6821, and ertr 6822. (Contributed by NM, 4-Jun-1995.) (Revised by Mario Carneiro, 2-Nov-2015.)
(𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅))
 
Theoremdfer2 6808* Alternate definition of equivalence predicate. (Contributed by NM, 3-Jan-1997.) (Revised by Mario Carneiro, 12-Aug-2015.)
(𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ ∀𝑥∀𝑦∀𝑧((𝑥𝑅𝑦 → 𝑦𝑅𝑥) ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧))))
 
Definitiondf-ec 6809 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 (i.e. when Er 𝑅; see dfer2 6808). In this case, 𝐴 is a representative (member) of the equivalence class [𝐴]𝑅, which contains all sets that are equivalent to 𝐴. Definition of [Enderton] p. 57 uses the notation [𝐴] (subscript) 𝑅, although we simply follow the brackets by 𝑅 since we don't have subscripted expressions. For an alternate definition, see dfec2 6810. (Contributed by NM, 23-Jul-1995.)
[𝐴]𝑅 = (𝑅 “ {𝐴})
 
Theoremdfec2 6810* Alternate definition of 𝑅-coset of 𝐴. Definition 34 of [Suppes] p. 81. (Contributed by NM, 3-Jan-1997.) (Proof shortened by Mario Carneiro, 9-Jul-2014.)
(𝐴 ∈ 𝑉 → [𝐴]𝑅 = {𝑦 ∣ 𝐴𝑅𝑦})
 
Theoremecexg 6811 An equivalence class modulo a set is a set. (Contributed by NM, 24-Jul-1995.)
(𝑅 ∈ 𝐵 → [𝐴]𝑅 ∈ V)
 
Theoremecexr 6812 An inhabited equivalence class implies the representative is a set. (Contributed by Mario Carneiro, 9-Jul-2014.)
(𝐴 ∈ [𝐵]𝑅 → 𝐵 ∈ V)
 
Definitiondf-qs 6813* Define quotient set. 𝑅 is usually an equivalence relation. Definition of [Enderton] p. 58. (Contributed by NM, 23-Jul-1995.)
(𝐴 / 𝑅) = {𝑦 ∣ ∃𝑥 ∈ 𝐴 𝑦 = [𝑥]𝑅}
 
Theoremereq1 6814 Equality theorem for equivalence predicate. (Contributed by NM, 4-Jun-1995.) (Revised by Mario Carneiro, 12-Aug-2015.)
(𝑅 = 𝑆 → (𝑅 Er 𝐴 ↔ 𝑆 Er 𝐴))
 
Theoremereq2 6815 Equality theorem for equivalence predicate. (Contributed by Mario Carneiro, 12-Aug-2015.)
(𝐴 = 𝐵 → (𝑅 Er 𝐴 ↔ 𝑅 Er 𝐵))
 
Theoremerrel 6816 An equivalence relation is a relation. (Contributed by Mario Carneiro, 12-Aug-2015.)
(𝑅 Er 𝐴 → Rel 𝑅)
 
Theoremerdm 6817 The domain of an equivalence relation. (Contributed by Mario Carneiro, 12-Aug-2015.)
(𝑅 Er 𝐴 → dom 𝑅 = 𝐴)
 
Theoremercl 6818 Elementhood in the field of an equivalence relation. (Contributed by Mario Carneiro, 12-Aug-2015.)
(𝜑 → 𝑅 Er 𝑋)    &   (𝜑 → 𝐴𝑅𝐵)    ⇒   (𝜑 → 𝐴 ∈ 𝑋)
 
Theoremersym 6819 An equivalence relation is symmetric. (Contributed by NM, 4-Jun-1995.) (Revised by Mario Carneiro, 12-Aug-2015.)
(𝜑 → 𝑅 Er 𝑋)    &   (𝜑 → 𝐴𝑅𝐵)    ⇒   (𝜑 → 𝐵𝑅𝐴)
 
Theoremercl2 6820 Elementhood in the field of an equivalence relation. (Contributed by Mario Carneiro, 12-Aug-2015.)
(𝜑 → 𝑅 Er 𝑋)    &   (𝜑 → 𝐴𝑅𝐵)    ⇒   (𝜑 → 𝐵 ∈ 𝑋)
 
Theoremersymb 6821 An equivalence relation is symmetric. (Contributed by NM, 30-Jul-1995.) (Revised by Mario Carneiro, 12-Aug-2015.)
(𝜑 → 𝑅 Er 𝑋)    ⇒   (𝜑 → (𝐴𝑅𝐵 ↔ 𝐵𝑅𝐴))
 
Theoremertr 6822 An equivalence relation is transitive. (Contributed by NM, 4-Jun-1995.) (Revised by Mario Carneiro, 12-Aug-2015.)
(𝜑 → 𝑅 Er 𝑋)    ⇒   (𝜑 → ((𝐴𝑅𝐵 ∧ 𝐵𝑅𝐶) → 𝐴𝑅𝐶))
 
Theoremertrd 6823 A transitivity relation for equivalences. (Contributed by Mario Carneiro, 9-Jul-2014.)
(𝜑 → 𝑅 Er 𝑋)    &   (𝜑 → 𝐴𝑅𝐵)    &   (𝜑 → 𝐵𝑅𝐶)    ⇒   (𝜑 → 𝐴𝑅𝐶)
 
Theoremertr2d 6824 A transitivity relation for equivalences. (Contributed by Mario Carneiro, 9-Jul-2014.)
(𝜑 → 𝑅 Er 𝑋)    &   (𝜑 → 𝐴𝑅𝐵)    &   (𝜑 → 𝐵𝑅𝐶)    ⇒   (𝜑 → 𝐶𝑅𝐴)
 
Theoremertr3d 6825 A transitivity relation for equivalences. (Contributed by Mario Carneiro, 9-Jul-2014.)
(𝜑 → 𝑅 Er 𝑋)    &   (𝜑 → 𝐵𝑅𝐴)    &   (𝜑 → 𝐵𝑅𝐶)    ⇒   (𝜑 → 𝐴𝑅𝐶)
 
Theoremertr4d 6826 A transitivity relation for equivalences. (Contributed by Mario Carneiro, 9-Jul-2014.)
(𝜑 → 𝑅 Er 𝑋)    &   (𝜑 → 𝐴𝑅𝐵)    &   (𝜑 → 𝐶𝑅𝐵)    ⇒   (𝜑 → 𝐴𝑅𝐶)
 
Theoremerref 6827 An equivalence relation is reflexive on its field. Compare Theorem 3M of [Enderton] p. 56. (Contributed by Mario Carneiro, 6-May-2013.) (Revised by Mario Carneiro, 12-Aug-2015.)
(𝜑 → 𝑅 Er 𝑋)    &   (𝜑 → 𝐴 ∈ 𝑋)    ⇒   (𝜑 → 𝐴𝑅𝐴)
 
Theoremercnv 6828 The converse of an equivalence relation is itself. (Contributed by Mario Carneiro, 12-Aug-2015.)
(𝑅 Er 𝐴 → ◡𝑅 = 𝑅)
 
Theoremerrn 6829 The range and domain of an equivalence relation are equal. (Contributed by Rodolfo Medina, 11-Oct-2010.) (Revised by Mario Carneiro, 12-Aug-2015.)
(𝑅 Er 𝐴 → ran 𝑅 = 𝐴)
 
Theoremerssxp 6830 An equivalence relation is a subset of the cartesian product of the field. (Contributed by Mario Carneiro, 12-Aug-2015.)
(𝑅 Er 𝐴 → 𝑅 ⊆ (𝐴 × 𝐴))
 
Theoremerex 6831 An equivalence relation is a set if its domain is a set. (Contributed by Rodolfo Medina, 15-Oct-2010.) (Proof shortened by Mario Carneiro, 12-Aug-2015.)
(𝑅 Er 𝐴 → (𝐴 ∈ 𝑉 → 𝑅 ∈ V))
 
Theoremerexb 6832 An equivalence relation is a set if and only if its domain is a set. (Contributed by Rodolfo Medina, 15-Oct-2010.) (Revised by Mario Carneiro, 12-Aug-2015.)
(𝑅 Er 𝐴 → (𝑅 ∈ V ↔ 𝐴 ∈ V))
 
Theoremiserd 6833* A reflexive, symmetric, transitive relation is an equivalence relation on its domain. (Contributed by Mario Carneiro, 9-Jul-2014.) (Revised by Mario Carneiro, 12-Aug-2015.)
(𝜑 → Rel 𝑅)    &   ((𝜑 ∧ 𝑥𝑅𝑦) → 𝑦𝑅𝑥)    &   ((𝜑 ∧ (𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧)) → 𝑥𝑅𝑧)    &   (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝑥𝑅𝑥))    ⇒   (𝜑 → 𝑅 Er 𝐴)
 
Theorembrdifun 6834 Evaluate the incomparability relation. (Contributed by Mario Carneiro, 9-Jul-2014.)
𝑅 = ((𝑋 × 𝑋) ∖ ( < ∪ ◡ < ))    ⇒   ((𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋) → (𝐴𝑅𝐵 ↔ ¬ (𝐴 < 𝐵 ∨ 𝐵 < 𝐴)))
 
Theoremswoer 6835* Incomparability under a strict weak partial order is an equivalence relation. (Contributed by Mario Carneiro, 9-Jul-2014.) (Revised by Mario Carneiro, 12-Aug-2015.)
𝑅 = ((𝑋 × 𝑋) ∖ ( < ∪ ◡ < ))    &   ((𝜑 ∧ (𝑦 ∈ 𝑋 ∧ 𝑧 ∈ 𝑋)) → (𝑦 < 𝑧 → ¬ 𝑧 < 𝑦))    &   ((𝜑 ∧ (𝑥 ∈ 𝑋 ∧ 𝑦 ∈ 𝑋 ∧ 𝑧 ∈ 𝑋)) → (𝑥 < 𝑦 → (𝑥 < 𝑧 ∨ 𝑧 < 𝑦)))    ⇒   (𝜑 → 𝑅 Er 𝑋)
 
Theoremswoord1 6836* The incomparability equivalence relation is compatible with the original order. (Contributed by Mario Carneiro, 31-Dec-2014.)
𝑅 = ((𝑋 × 𝑋) ∖ ( < ∪ ◡ < ))    &   ((𝜑 ∧ (𝑦 ∈ 𝑋 ∧ 𝑧 ∈ 𝑋)) → (𝑦 < 𝑧 → ¬ 𝑧 < 𝑦))    &   ((𝜑 ∧ (𝑥 ∈ 𝑋 ∧ 𝑦 ∈ 𝑋 ∧ 𝑧 ∈ 𝑋)) → (𝑥 < 𝑦 → (𝑥 < 𝑧 ∨ 𝑧 < 𝑦)))    &   (𝜑 → 𝐵 ∈ 𝑋)    &   (𝜑 → 𝐶 ∈ 𝑋)    &   (𝜑 → 𝐴𝑅𝐵)    ⇒   (𝜑 → (𝐴 < 𝐶 ↔ 𝐵 < 𝐶))
 
Theoremswoord2 6837* The incomparability equivalence relation is compatible with the original order. (Contributed by Mario Carneiro, 31-Dec-2014.)
𝑅 = ((𝑋 × 𝑋) ∖ ( < ∪ ◡ < ))    &   ((𝜑 ∧ (𝑦 ∈ 𝑋 ∧ 𝑧 ∈ 𝑋)) → (𝑦 < 𝑧 → ¬ 𝑧 < 𝑦))    &   ((𝜑 ∧ (𝑥 ∈ 𝑋 ∧ 𝑦 ∈ 𝑋 ∧ 𝑧 ∈ 𝑋)) → (𝑥 < 𝑦 → (𝑥 < 𝑧 ∨ 𝑧 < 𝑦)))    &   (𝜑 → 𝐵 ∈ 𝑋)    &   (𝜑 → 𝐶 ∈ 𝑋)    &   (𝜑 → 𝐴𝑅𝐵)    ⇒   (𝜑 → (𝐶 < 𝐴 ↔ 𝐶 < 𝐵))
 
Theoremeqerlem 6838* Lemma for eqer 6839. (Contributed by NM, 17-Mar-2008.) (Proof shortened by Mario Carneiro, 6-Dec-2016.)
(𝑥 = 𝑦 → 𝐴 = 𝐵)    &   𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝐴 = 𝐵}    ⇒   (𝑧𝑅𝑤 ↔ ⦋𝑧 / 𝑥⦌𝐴 = ⦋𝑤 / 𝑥⦌𝐴)
 
Theoremeqer 6839* Equivalence relation involving equality of dependent classes 𝐴(𝑥) and 𝐵(𝑦). (Contributed by NM, 17-Mar-2008.) (Revised by Mario Carneiro, 12-Aug-2015.)
(𝑥 = 𝑦 → 𝐴 = 𝐵)    &   𝑅 = {⟨𝑥, 𝑦⟩ ∣ 𝐴 = 𝐵}    ⇒   𝑅 Er V
 
Theoremider 6840 The identity relation is an equivalence relation. (Contributed by NM, 10-May-1998.) (Proof shortened by Andrew Salmon, 22-Oct-2011.) (Proof shortened by Mario Carneiro, 9-Jul-2014.)
I Er V
 
Theorem0er 6841 The empty set is an equivalence relation on the empty set. (Contributed by Mario Carneiro, 5-Sep-2015.)
∅ Er ∅
 
Theoremeceq1 6842 Equality theorem for equivalence class. (Contributed by NM, 23-Jul-1995.)
(𝐴 = 𝐵 → [𝐴]𝐶 = [𝐵]𝐶)
 
Theoremeceq1d 6843 Equality theorem for equivalence class (deduction form). (Contributed by Jim Kingdon, 31-Dec-2019.)
(𝜑 → 𝐴 = 𝐵)    ⇒   (𝜑 → [𝐴]𝐶 = [𝐵]𝐶)
 
Theoremeceq2 6844 Equality theorem for equivalence class. (Contributed by NM, 23-Jul-1995.)
(𝐴 = 𝐵 → [𝐶]𝐴 = [𝐶]𝐵)
 
Theoremeceq2i 6845 Equality theorem for the 𝐴-coset and 𝐵-coset of 𝐶, inference version. (Contributed by Peter Mazsa, 11-May-2021.)
𝐴 = 𝐵    ⇒   [𝐶]𝐴 = [𝐶]𝐵
 
Theoremeceq2d 6846 Equality theorem for the 𝐴-coset and 𝐵-coset of 𝐶, deduction version. (Contributed by Peter Mazsa, 23-Apr-2021.)
(𝜑 → 𝐴 = 𝐵)    ⇒   (𝜑 → [𝐶]𝐴 = [𝐶]𝐵)
 
Theoremelecg 6847 Membership in an equivalence class. Theorem 72 of [Suppes] p. 82. (Contributed by Mario Carneiro, 9-Jul-2014.)
((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (𝐴 ∈ [𝐵]𝑅 ↔ 𝐵𝑅𝐴))
 
Theoremelec 6848 Membership in an equivalence class. Theorem 72 of [Suppes] p. 82. (Contributed by NM, 23-Jul-1995.)
𝐴 ∈ V    &   𝐵 ∈ V    ⇒   (𝐴 ∈ [𝐵]𝑅 ↔ 𝐵𝑅𝐴)
 
Theoremrelelec 6849 Membership in an equivalence class when 𝑅 is a relation. (Contributed by Mario Carneiro, 11-Sep-2015.)
(Rel 𝑅 → (𝐴 ∈ [𝐵]𝑅 ↔ 𝐵𝑅𝐴))
 
Theoremecss 6850 An equivalence class is a subset of the domain. (Contributed by NM, 6-Aug-1995.) (Revised by Mario Carneiro, 12-Aug-2015.)
(𝜑 → 𝑅 Er 𝑋)    ⇒   (𝜑 → [𝐴]𝑅 ⊆ 𝑋)
 
Theoremecdmn0m 6851* A representative of an inhabited equivalence class belongs to the domain of the equivalence relation. (Contributed by Jim Kingdon, 21-Aug-2019.)
(𝐴 ∈ dom 𝑅 ↔ ∃𝑥 𝑥 ∈ [𝐴]𝑅)
 
Theoremereldm 6852 Equality of equivalence classes implies equivalence of domain membership. (Contributed by NM, 28-Jan-1996.) (Revised by Mario Carneiro, 12-Aug-2015.)
(𝜑 → 𝑅 Er 𝑋)    &   (𝜑 → [𝐴]𝑅 = [𝐵]𝑅)    ⇒   (𝜑 → (𝐴 ∈ 𝑋 ↔ 𝐵 ∈ 𝑋))
 
Theoremerth 6853 Basic property of equivalence relations. Theorem 73 of [Suppes] p. 82. (Contributed by NM, 23-Jul-1995.) (Revised by Mario Carneiro, 6-Jul-2015.)
(𝜑 → 𝑅 Er 𝑋)    &   (𝜑 → 𝐴 ∈ 𝑋)    ⇒   (𝜑 → (𝐴𝑅𝐵 ↔ [𝐴]𝑅 = [𝐵]𝑅))
 
Theoremerth2 6854 Basic property of equivalence relations. Compare Theorem 73 of [Suppes] p. 82. Assumes membership of the second argument in the domain. (Contributed by NM, 30-Jul-1995.) (Revised by Mario Carneiro, 6-Jul-2015.)
(𝜑 → 𝑅 Er 𝑋)    &   (𝜑 → 𝐵 ∈ 𝑋)    ⇒   (𝜑 → (𝐴𝑅𝐵 ↔ [𝐴]𝑅 = [𝐵]𝑅))
 
Theoremerthi 6855 Basic property of equivalence relations. Part of Lemma 3N of [Enderton] p. 57. (Contributed by NM, 30-Jul-1995.) (Revised by Mario Carneiro, 9-Jul-2014.)
(𝜑 → 𝑅 Er 𝑋)    &   (𝜑 → 𝐴𝑅𝐵)    ⇒   (𝜑 → [𝐴]𝑅 = [𝐵]𝑅)
 
Theoremecidsn 6856 An equivalence class modulo the identity relation is a singleton. (Contributed by NM, 24-Oct-2004.)
[𝐴] I = {𝐴}
 
Theoremqseq1 6857 Equality theorem for quotient set. (Contributed by NM, 23-Jul-1995.)
(𝐴 = 𝐵 → (𝐴 / 𝐶) = (𝐵 / 𝐶))
 
Theoremqseq2 6858 Equality theorem for quotient set. (Contributed by NM, 23-Jul-1995.)
(𝐴 = 𝐵 → (𝐶 / 𝐴) = (𝐶 / 𝐵))
 
Theoremelqsg 6859* Closed form of elqs 6860. (Contributed by Rodolfo Medina, 12-Oct-2010.)
(𝐵 ∈ 𝑉 → (𝐵 ∈ (𝐴 / 𝑅) ↔ ∃𝑥 ∈ 𝐴 𝐵 = [𝑥]𝑅))
 
Theoremelqs 6860* Membership in a quotient set. (Contributed by NM, 23-Jul-1995.)
𝐵 ∈ V    ⇒   (𝐵 ∈ (𝐴 / 𝑅) ↔ ∃𝑥 ∈ 𝐴 𝐵 = [𝑥]𝑅)
 
Theoremelqsi 6861* Membership in a quotient set. (Contributed by NM, 23-Jul-1995.)
(𝐵 ∈ (𝐴 / 𝑅) → ∃𝑥 ∈ 𝐴 𝐵 = [𝑥]𝑅)
 
Theoremecelqsg 6862 Membership of an equivalence class in a quotient set. (Contributed by Jeff Madsen, 10-Jun-2010.) (Revised by Mario Carneiro, 9-Jul-2014.)
((𝑅 ∈ 𝑉 ∧ 𝐵 ∈ 𝐴) → [𝐵]𝑅 ∈ (𝐴 / 𝑅))
 
Theoremecelqsi 6863 Membership of an equivalence class in a quotient set. (Contributed by NM, 25-Jul-1995.) (Revised by Mario Carneiro, 9-Jul-2014.)
𝑅 ∈ V    ⇒   (𝐵 ∈ 𝐴 → [𝐵]𝑅 ∈ (𝐴 / 𝑅))
 
Theoremecopqsi 6864 "Closure" law for equivalence class of ordered pairs. (Contributed by NM, 25-Mar-1996.)
𝑅 ∈ V    &   𝑆 = ((𝐴 × 𝐴) / 𝑅)    ⇒   ((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ 𝐴) → [⟨𝐵, 𝐶⟩]𝑅 ∈ 𝑆)
 
Theoremqsexg 6865 A quotient set exists. (Contributed by FL, 19-May-2007.) (Revised by Mario Carneiro, 9-Jul-2014.)
(𝐴 ∈ 𝑉 → (𝐴 / 𝑅) ∈ V)
 
Theoremqsex 6866 A quotient set exists. (Contributed by NM, 14-Aug-1995.)
𝐴 ∈ V    ⇒   (𝐴 / 𝑅) ∈ V
 
Theoremuniqs 6867 The union of a quotient set. (Contributed by NM, 9-Dec-2008.)
(𝑅 ∈ 𝑉 → ∪ (𝐴 / 𝑅) = (𝑅 “ 𝐴))
 
Theoremqsss 6868 A quotient set is a set of subsets of the base set. (Contributed by Mario Carneiro, 9-Jul-2014.) (Revised by Mario Carneiro, 12-Aug-2015.)
(𝜑 → 𝑅 Er 𝐴)    ⇒   (𝜑 → (𝐴 / 𝑅) ⊆ 𝒫 𝐴)
 
Theoremuniqs2 6869 The union of a quotient set. (Contributed by Mario Carneiro, 11-Jul-2014.)
(𝜑 → 𝑅 Er 𝐴)    &   (𝜑 → 𝑅 ∈ 𝑉)    ⇒   (𝜑 → ∪ (𝐴 / 𝑅) = 𝐴)
 
Theoremsnec 6870 The singleton of an equivalence class. (Contributed by NM, 29-Jan-1999.) (Revised by Mario Carneiro, 9-Jul-2014.)
𝐴 ∈ V    ⇒   {[𝐴]𝑅} = ({𝐴} / 𝑅)
 
Theoremecqs 6871 Equivalence class in terms of quotient set. (Contributed by NM, 29-Jan-1999.)
𝑅 ∈ V    ⇒   [𝐴]𝑅 = ∪ ({𝐴} / 𝑅)
 
Theoremecid 6872 A set is equal to its converse epsilon coset. (Note: converse epsilon is not an equivalence relation.) (Contributed by NM, 13-Aug-1995.) (Revised by Mario Carneiro, 9-Jul-2014.)
𝐴 ∈ V    ⇒   [𝐴]◡ E = 𝐴
 
Theoremecidg 6873 A set is equal to its converse epsilon coset. (Note: converse epsilon is not an equivalence relation.) (Contributed by Jim Kingdon, 8-Jan-2020.)
(𝐴 ∈ 𝑉 → [𝐴]◡ E = 𝐴)
 
Theoremqsid 6874 A set is equal to its quotient set mod converse epsilon. (Note: converse epsilon is not an equivalence relation.) (Contributed by NM, 13-Aug-1995.) (Revised by Mario Carneiro, 9-Jul-2014.)
(𝐴 / ◡ E ) = 𝐴
 
Theoremectocld 6875* Implicit substitution of class for equivalence class. (Contributed by Mario Carneiro, 9-Jul-2014.)
𝑆 = (𝐵 / 𝑅)    &   ([𝑥]𝑅 = 𝐴 → (𝜑 ↔ 𝜓))    &   ((𝜒 ∧ 𝑥 ∈ 𝐵) → 𝜑)    ⇒   ((𝜒 ∧ 𝐴 ∈ 𝑆) → 𝜓)
 
Theoremectocl 6876* Implicit substitution of class for equivalence class. (Contributed by NM, 23-Jul-1995.) (Revised by Mario Carneiro, 9-Jul-2014.)
𝑆 = (𝐵 / 𝑅)    &   ([𝑥]𝑅 = 𝐴 → (𝜑 ↔ 𝜓))    &   (𝑥 ∈ 𝐵 → 𝜑)    ⇒   (𝐴 ∈ 𝑆 → 𝜓)
 
Theoremelqsn0m 6877* An element of a quotient set is inhabited. (Contributed by Jim Kingdon, 21-Aug-2019.)
((dom 𝑅 = 𝐴 ∧ 𝐵 ∈ (𝐴 / 𝑅)) → ∃𝑥 𝑥 ∈ 𝐵)
 
Theoremelqsn0 6878 A quotient set doesn't contain the empty set. (Contributed by NM, 24-Aug-1995.)
((dom 𝑅 = 𝐴 ∧ 𝐵 ∈ (𝐴 / 𝑅)) → 𝐵 ≠ ∅)
 
Theoremecelqsdm 6879 Membership of an equivalence class in a quotient set. (Contributed by NM, 30-Jul-1995.)
((dom 𝑅 = 𝐴 ∧ [𝐵]𝑅 ∈ (𝐴 / 𝑅)) → 𝐵 ∈ 𝐴)
 
Theoremxpider 6880 A square Cartesian product is an equivalence relation (in general it's not a poset). (Contributed by FL, 31-Jul-2009.) (Revised by Mario Carneiro, 12-Aug-2015.)
(𝐴 × 𝐴) Er 𝐴
 
Theoremiinerm 6881* The intersection of a nonempty family of equivalence relations is an equivalence relation. (Contributed by Mario Carneiro, 27-Sep-2015.)
((∃𝑦 𝑦 ∈ 𝐴 ∧ ∀𝑥 ∈ 𝐴 𝑅 Er 𝐵) → ∩ 𝑥 ∈ 𝐴 𝑅 Er 𝐵)
 
Theoremriinerm 6882* The relative intersection of a family of equivalence relations is an equivalence relation. (Contributed by Mario Carneiro, 27-Sep-2015.)
((∃𝑦 𝑦 ∈ 𝐴 ∧ ∀𝑥 ∈ 𝐴 𝑅 Er 𝐵) → ((𝐵 × 𝐵) ∩ ∩ 𝑥 ∈ 𝐴 𝑅) Er 𝐵)
 
Theoremerinxp 6883 A restricted equivalence relation is an equivalence relation. (Contributed by Mario Carneiro, 10-Jul-2015.) (Revised by Mario Carneiro, 12-Aug-2015.)
(𝜑 → 𝑅 Er 𝐴)    &   (𝜑 → 𝐵 ⊆ 𝐴)    ⇒   (𝜑 → (𝑅 ∩ (𝐵 × 𝐵)) Er 𝐵)
 
Theoremecinxp 6884 Restrict the relation in an equivalence class to a base set. (Contributed by Mario Carneiro, 10-Jul-2015.)
(((𝑅 “ 𝐴) ⊆ 𝐴 ∧ 𝐵 ∈ 𝐴) → [𝐵]𝑅 = [𝐵](𝑅 ∩ (𝐴 × 𝐴)))
 
Theoremqsinxp 6885 Restrict the equivalence relation in a quotient set to the base set. (Contributed by Mario Carneiro, 23-Feb-2015.)
((𝑅 “ 𝐴) ⊆ 𝐴 → (𝐴 / 𝑅) = (𝐴 / (𝑅 ∩ (𝐴 × 𝐴))))
 
Theoremqsel 6886 If an element of a quotient set contains a given element, it is equal to the equivalence class of the element. (Contributed by Mario Carneiro, 12-Aug-2015.)
((𝑅 Er 𝑋 ∧ 𝐵 ∈ (𝐴 / 𝑅) ∧ 𝐶 ∈ 𝐵) → 𝐵 = [𝐶]𝑅)
 
Theoremqliftlem 6887* 𝐹, a function lift, is a subset of 𝑅 × 𝑆. (Contributed by Mario Carneiro, 23-Dec-2016.)
𝐹 = ran (𝑥 ∈ 𝑋 ↦ ⟨[𝑥]𝑅, 𝐴⟩)    &   ((𝜑 ∧ 𝑥 ∈ 𝑋) → 𝐴 ∈ 𝑌)    &   (𝜑 → 𝑅 Er 𝑋)    &   (𝜑 → 𝑋 ∈ V)    ⇒   ((𝜑 ∧ 𝑥 ∈ 𝑋) → [𝑥]𝑅 ∈ (𝑋 / 𝑅))
 
Theoremqliftrel 6888* 𝐹, a function lift, is a subset of 𝑅 × 𝑆. (Contributed by Mario Carneiro, 23-Dec-2016.)
𝐹 = ran (𝑥 ∈ 𝑋 ↦ ⟨[𝑥]𝑅, 𝐴⟩)    &   ((𝜑 ∧ 𝑥 ∈ 𝑋) → 𝐴 ∈ 𝑌)    &   (𝜑 → 𝑅 Er 𝑋)    &   (𝜑 → 𝑋 ∈ V)    ⇒   (𝜑 → 𝐹 ⊆ ((𝑋 / 𝑅) × 𝑌))
 
Theoremqliftel 6889* Elementhood in the relation 𝐹. (Contributed by Mario Carneiro, 23-Dec-2016.)
𝐹 = ran (𝑥 ∈ 𝑋 ↦ ⟨[𝑥]𝑅, 𝐴⟩)    &   ((𝜑 ∧ 𝑥 ∈ 𝑋) → 𝐴 ∈ 𝑌)    &   (𝜑 → 𝑅 Er 𝑋)    &   (𝜑 → 𝑋 ∈ V)    ⇒   (𝜑 → ([𝐶]𝑅𝐹𝐷 ↔ ∃𝑥 ∈ 𝑋 (𝐶𝑅𝑥 ∧ 𝐷 = 𝐴)))
 
Theoremqliftel1 6890* Elementhood in the relation 𝐹. (Contributed by Mario Carneiro, 23-Dec-2016.)
𝐹 = ran (𝑥 ∈ 𝑋 ↦ ⟨[𝑥]𝑅, 𝐴⟩)    &   ((𝜑 ∧ 𝑥 ∈ 𝑋) → 𝐴 ∈ 𝑌)    &   (𝜑 → 𝑅 Er 𝑋)    &   (𝜑 → 𝑋 ∈ V)    ⇒   ((𝜑 ∧ 𝑥 ∈ 𝑋) → [𝑥]𝑅𝐹𝐴)
 
Theoremqliftfun 6891* The function 𝐹 is the unique function defined by 𝐹‘[𝑥] = 𝐴, provided that the well-definedness condition holds. (Contributed by Mario Carneiro, 23-Dec-2016.)
𝐹 = ran (𝑥 ∈ 𝑋 ↦ ⟨[𝑥]𝑅, 𝐴⟩)    &   ((𝜑 ∧ 𝑥 ∈ 𝑋) → 𝐴 ∈ 𝑌)    &   (𝜑 → 𝑅 Er 𝑋)    &   (𝜑 → 𝑋 ∈ V)    &   (𝑥 = 𝑦 → 𝐴 = 𝐵)    ⇒   (𝜑 → (Fun 𝐹 ↔ ∀𝑥∀𝑦(𝑥𝑅𝑦 → 𝐴 = 𝐵)))
 
Theoremqliftfund 6892* The function 𝐹 is the unique function defined by 𝐹‘[𝑥] = 𝐴, provided that the well-definedness condition holds. (Contributed by Mario Carneiro, 23-Dec-2016.)
𝐹 = ran (𝑥 ∈ 𝑋 ↦ ⟨[𝑥]𝑅, 𝐴⟩)    &   ((𝜑 ∧ 𝑥 ∈ 𝑋) → 𝐴 ∈ 𝑌)    &   (𝜑 → 𝑅 Er 𝑋)    &   (𝜑 → 𝑋 ∈ V)    &   (𝑥 = 𝑦 → 𝐴 = 𝐵)    &   ((𝜑 ∧ 𝑥𝑅𝑦) → 𝐴 = 𝐵)    ⇒   (𝜑 → Fun 𝐹)
 
Theoremqliftfuns 6893* The function 𝐹 is the unique function defined by 𝐹‘[𝑥] = 𝐴, provided that the well-definedness condition holds. (Contributed by Mario Carneiro, 23-Dec-2016.)
𝐹 = ran (𝑥 ∈ 𝑋 ↦ ⟨[𝑥]𝑅, 𝐴⟩)    &   ((𝜑 ∧ 𝑥 ∈ 𝑋) → 𝐴 ∈ 𝑌)    &   (𝜑 → 𝑅 Er 𝑋)    &   (𝜑 → 𝑋 ∈ V)    ⇒   (𝜑 → (Fun 𝐹 ↔ ∀𝑦∀𝑧(𝑦𝑅𝑧 → ⦋𝑦 / 𝑥⦌𝐴 = ⦋𝑧 / 𝑥⦌𝐴)))
 
Theoremqliftf 6894* The domain and codomain of the function 𝐹. (Contributed by Mario Carneiro, 23-Dec-2016.)
𝐹 = ran (𝑥 ∈ 𝑋 ↦ ⟨[𝑥]𝑅, 𝐴⟩)    &   ((𝜑 ∧ 𝑥 ∈ 𝑋) → 𝐴 ∈ 𝑌)    &   (𝜑 → 𝑅 Er 𝑋)    &   (𝜑 → 𝑋 ∈ V)    ⇒   (𝜑 → (Fun 𝐹 ↔ 𝐹:(𝑋 / 𝑅)⟶𝑌))
 
Theoremqliftval 6895* The value of the function 𝐹. (Contributed by Mario Carneiro, 23-Dec-2016.)
𝐹 = ran (𝑥 ∈ 𝑋 ↦ ⟨[𝑥]𝑅, 𝐴⟩)    &   ((𝜑 ∧ 𝑥 ∈ 𝑋) → 𝐴 ∈ 𝑌)    &   (𝜑 → 𝑅 Er 𝑋)    &   (𝜑 → 𝑋 ∈ V)    &   (𝑥 = 𝐶 → 𝐴 = 𝐵)    &   (𝜑 → Fun 𝐹)    ⇒   ((𝜑 ∧ 𝐶 ∈ 𝑋) → (𝐹‘[𝐶]𝑅) = 𝐵)
 
Theoremecoptocl 6896* Implicit substitution of class for equivalence class of ordered pair. (Contributed by NM, 23-Jul-1995.)
𝑆 = ((𝐵 × 𝐶) / 𝑅)    &   ([⟨𝑥, 𝑦⟩]𝑅 = 𝐴 → (𝜑 ↔ 𝜓))    &   ((𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶) → 𝜑)    ⇒   (𝐴 ∈ 𝑆 → 𝜓)
 
Theorem2ecoptocl 6897* Implicit substitution of classes for equivalence classes of ordered pairs. (Contributed by NM, 23-Jul-1995.)
𝑆 = ((𝐶 × 𝐷) / 𝑅)    &   ([⟨𝑥, 𝑦⟩]𝑅 = 𝐴 → (𝜑 ↔ 𝜓))    &   ([⟨𝑧, 𝑤⟩]𝑅 = 𝐵 → (𝜓 ↔ 𝜒))    &   (((𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐷) ∧ (𝑧 ∈ 𝐶 ∧ 𝑤 ∈ 𝐷)) → 𝜑)    ⇒   ((𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆) → 𝜒)
 
Theorem3ecoptocl 6898* Implicit substitution of classes for equivalence classes of ordered pairs. (Contributed by NM, 9-Aug-1995.)
𝑆 = ((𝐷 × 𝐷) / 𝑅)    &   ([⟨𝑥, 𝑦⟩]𝑅 = 𝐴 → (𝜑 ↔ 𝜓))    &   ([⟨𝑧, 𝑤⟩]𝑅 = 𝐵 → (𝜓 ↔ 𝜒))    &   ([⟨𝑣, 𝑢⟩]𝑅 = 𝐶 → (𝜒 ↔ 𝜃))    &   (((𝑥 ∈ 𝐷 ∧ 𝑦 ∈ 𝐷) ∧ (𝑧 ∈ 𝐷 ∧ 𝑤 ∈ 𝐷) ∧ (𝑣 ∈ 𝐷 ∧ 𝑢 ∈ 𝐷)) → 𝜑)    ⇒   ((𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆 ∧ 𝐶 ∈ 𝑆) → 𝜃)
 
Theorembrecop 6899* Binary relation on a quotient set. Lemma for real number construction. (Contributed by NM, 29-Jan-1996.)
∼ ∈ V    &    ∼ Er (𝐺 × 𝐺)    &   𝐻 = ((𝐺 × 𝐺) / ∼ )    &    ≤ = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ 𝐻 ∧ 𝑦 ∈ 𝐻) ∧ ∃𝑧∃𝑤∃𝑣∃𝑢((𝑥 = [⟨𝑧, 𝑤⟩] ∼ ∧ 𝑦 = [⟨𝑣, 𝑢⟩] ∼ ) ∧ 𝜑))}    &   ((((𝑧 ∈ 𝐺 ∧ 𝑤 ∈ 𝐺) ∧ (𝐴 ∈ 𝐺 ∧ 𝐵 ∈ 𝐺)) ∧ ((𝑣 ∈ 𝐺 ∧ 𝑢 ∈ 𝐺) ∧ (𝐶 ∈ 𝐺 ∧ 𝐷 ∈ 𝐺))) → (([⟨𝑧, 𝑤⟩] ∼ = [⟨𝐴, 𝐵⟩] ∼ ∧ [⟨𝑣, 𝑢⟩] ∼ = [⟨𝐶, 𝐷⟩] ∼ ) → (𝜑 ↔ 𝜓)))    ⇒   (((𝐴 ∈ 𝐺 ∧ 𝐵 ∈ 𝐺) ∧ (𝐶 ∈ 𝐺 ∧ 𝐷 ∈ 𝐺)) → ([⟨𝐴, 𝐵⟩] ∼ ≤ [⟨𝐶, 𝐷⟩] ∼ ↔ 𝜓))
 
Theoremeroveu 6900* Lemma for eroprf 6902. (Contributed by Jeff Madsen, 10-Jun-2010.) (Revised by Mario Carneiro, 9-Jul-2014.)
𝐽 = (𝐴 / 𝑅)    &   𝐾 = (𝐵 / 𝑆)    &   (𝜑 → 𝑇 ∈ 𝑍)    &   (𝜑 → 𝑅 Er 𝑈)    &   (𝜑 → 𝑆 Er 𝑉)    &   (𝜑 → 𝑇 Er 𝑊)    &   (𝜑 → 𝐴 ⊆ 𝑈)    &   (𝜑 → 𝐵 ⊆ 𝑉)    &   (𝜑 → 𝐶 ⊆ 𝑊)    &   (𝜑 → + :(𝐴 × 𝐵)⟶𝐶)    &   ((𝜑 ∧ ((𝑟 ∈ 𝐴 ∧ 𝑠 ∈ 𝐴) ∧ (𝑡 ∈ 𝐵 ∧ 𝑢 ∈ 𝐵))) → ((𝑟𝑅𝑠 ∧ 𝑡𝑆𝑢) → (𝑟 + 𝑡)𝑇(𝑠 + 𝑢)))    ⇒   ((𝜑 ∧ (𝑋 ∈ 𝐽 ∧ 𝑌 ∈ 𝐾)) → ∃!𝑧∃𝑝 ∈ 𝐴 ∃𝑞 ∈ 𝐵 ((𝑋 = [𝑝]𝑅 ∧ 𝑌 = [𝑞]𝑆) ∧ 𝑧 = [(𝑝 + 𝑞)]𝑇))
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