Theorem erdisj 8758

Description: Equivalence classes do not overlap. In other words, two equivalence classes are either equal or disjoint. Theorem 74 of [Suppes] p. 83. (Contributed by NM, 15-Jun-2004.) (Revised by Mario Carneiro, 9-Jul-2014.)

Assertion

Ref	Expression
erdisj	⊢ (𝑅 Er 𝑋 → ([𝐴]𝑅 = [𝐵]𝑅 ∨ ([𝐴]𝑅 ∩ [𝐵]𝑅) = ∅))

Proof of Theorem erdisj

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		neq0 4306	. . . 4 ⊢ (¬ ([𝐴]𝑅 ∩ [𝐵]𝑅) = ∅ ↔ ∃𝑥 𝑥 ∈ ([𝐴]𝑅 ∩ [𝐵]𝑅))
2		simpl 488	. . . . . . 7 ⊢ ((𝑅 Er 𝑋 ∧ 𝑥 ∈ ([𝐴]𝑅 ∩ [𝐵]𝑅)) → 𝑅 Er 𝑋)
3		elinel1 4154	. . . . . . . . . 10 ⊢ (𝑥 ∈ ([𝐴]𝑅 ∩ [𝐵]𝑅) → 𝑥 ∈ [𝐴]𝑅)
4	3	adantl 487	. . . . . . . . 9 ⊢ ((𝑅 Er 𝑋 ∧ 𝑥 ∈ ([𝐴]𝑅 ∩ [𝐵]𝑅)) → 𝑥 ∈ [𝐴]𝑅)
5		vex 3461	. . . . . . . . . 10 ⊢ 𝑥 ∈ V
6		ecexr 8705	. . . . . . . . . . 11 ⊢ (𝑥 ∈ [𝐴]𝑅 → 𝐴 ∈ V)
7	4, 6	syl 18	. . . . . . . . . 10 ⊢ ((𝑅 Er 𝑋 ∧ 𝑥 ∈ ([𝐴]𝑅 ∩ [𝐵]𝑅)) → 𝐴 ∈ V)
8		elecg 8745	. . . . . . . . . 10 ⊢ ((𝑥 ∈ V ∧ 𝐴 ∈ V) → (𝑥 ∈ [𝐴]𝑅 ↔ 𝐴𝑅𝑥))
9	5, 7, 8	sylancr 599	. . . . . . . . 9 ⊢ ((𝑅 Er 𝑋 ∧ 𝑥 ∈ ([𝐴]𝑅 ∩ [𝐵]𝑅)) → (𝑥 ∈ [𝐴]𝑅 ↔ 𝐴𝑅𝑥))
10	4, 9	mpbid 235	. . . . . . . 8 ⊢ ((𝑅 Er 𝑋 ∧ 𝑥 ∈ ([𝐴]𝑅 ∩ [𝐵]𝑅)) → 𝐴𝑅𝑥)
11		elinel2 4155	. . . . . . . . . 10 ⊢ (𝑥 ∈ ([𝐴]𝑅 ∩ [𝐵]𝑅) → 𝑥 ∈ [𝐵]𝑅)
12	11	adantl 487	. . . . . . . . 9 ⊢ ((𝑅 Er 𝑋 ∧ 𝑥 ∈ ([𝐴]𝑅 ∩ [𝐵]𝑅)) → 𝑥 ∈ [𝐵]𝑅)
13		ecexr 8705	. . . . . . . . . . 11 ⊢ (𝑥 ∈ [𝐵]𝑅 → 𝐵 ∈ V)
14	12, 13	syl 18	. . . . . . . . . 10 ⊢ ((𝑅 Er 𝑋 ∧ 𝑥 ∈ ([𝐴]𝑅 ∩ [𝐵]𝑅)) → 𝐵 ∈ V)
15		elecg 8745	. . . . . . . . . 10 ⊢ ((𝑥 ∈ V ∧ 𝐵 ∈ V) → (𝑥 ∈ [𝐵]𝑅 ↔ 𝐵𝑅𝑥))
16	5, 14, 15	sylancr 599	. . . . . . . . 9 ⊢ ((𝑅 Er 𝑋 ∧ 𝑥 ∈ ([𝐴]𝑅 ∩ [𝐵]𝑅)) → (𝑥 ∈ [𝐵]𝑅 ↔ 𝐵𝑅𝑥))
17	12, 16	mpbid 235	. . . . . . . 8 ⊢ ((𝑅 Er 𝑋 ∧ 𝑥 ∈ ([𝐴]𝑅 ∩ [𝐵]𝑅)) → 𝐵𝑅𝑥)
18	2, 10, 17	ertr4d 8720	. . . . . . 7 ⊢ ((𝑅 Er 𝑋 ∧ 𝑥 ∈ ([𝐴]𝑅 ∩ [𝐵]𝑅)) → 𝐴𝑅𝐵)
19	2, 18	erthi 8757	. . . . . 6 ⊢ ((𝑅 Er 𝑋 ∧ 𝑥 ∈ ([𝐴]𝑅 ∩ [𝐵]𝑅)) → [𝐴]𝑅 = [𝐵]𝑅)
20	19	ex 418	. . . . 5 ⊢ (𝑅 Er 𝑋 → (𝑥 ∈ ([𝐴]𝑅 ∩ [𝐵]𝑅) → [𝐴]𝑅 = [𝐵]𝑅))
21	20	exlimdv 1966	. . . 4 ⊢ (𝑅 Er 𝑋 → (∃𝑥 𝑥 ∈ ([𝐴]𝑅 ∩ [𝐵]𝑅) → [𝐴]𝑅 = [𝐵]𝑅))
22	1, 21	biimtrid 245	. . 3 ⊢ (𝑅 Er 𝑋 → (¬ ([𝐴]𝑅 ∩ [𝐵]𝑅) = ∅ → [𝐴]𝑅 = [𝐵]𝑅))
23	22	orrd 877	. 2 ⊢ (𝑅 Er 𝑋 → (([𝐴]𝑅 ∩ [𝐵]𝑅) = ∅ ∨ [𝐴]𝑅 = [𝐵]𝑅))
24	23	orcomd 885	1 ⊢ (𝑅 Er 𝑋 → ([𝐴]𝑅 = [𝐵]𝑅 ∨ ([𝐴]𝑅 ∩ [𝐵]𝑅) = ∅))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401   ∨ wo 861   = wceq 1570  ∃wex 1812   ∈ wcel 2146  Vcvv 3457   ∩ cin 3905  ∅c0 4286   class class class wbr 5111   Er wer 8697  [cec 8698

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737  ax-sep 5259  ax-pr 5406

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-ne 2961  df-ral 3082  df-rex 3092  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-nul 4287  df-if 4490  df-sn 4592  df-pr 4594  df-op 4598  df-br 5112  df-opab 5176  df-xp 5669  df-rel 5670  df-cnv 5671  df-co 5672  df-dm 5673  df-rn 5674  df-res 5675  df-ima 5676  df-er 8700  df-ec 8702

This theorem is used by:  qsdisj  8798