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Theorem ereq1 6049
Description: Equality theorem for equivalence predicate. (Contributed by NM, 4-Jun-1995.) (Revised by Mario Carneiro, 12-Aug-2015.)
Assertion
Ref Expression
ereq1 (𝑅 = 𝑆 → (𝑅 Er A𝑆 Er A))

Proof of Theorem ereq1
StepHypRef Expression
1 releq 4365 . . 3 (𝑅 = 𝑆 → (Rel 𝑅 ↔ Rel 𝑆))
2 dmeq 4478 . . . 4 (𝑅 = 𝑆 → dom 𝑅 = dom 𝑆)
32eqeq1d 2045 . . 3 (𝑅 = 𝑆 → (dom 𝑅 = A ↔ dom 𝑆 = A))
4 cnveq 4452 . . . . . 6 (𝑅 = 𝑆𝑅 = 𝑆)
5 coeq1 4436 . . . . . . 7 (𝑅 = 𝑆 → (𝑅𝑅) = (𝑆𝑅))
6 coeq2 4437 . . . . . . 7 (𝑅 = 𝑆 → (𝑆𝑅) = (𝑆𝑆))
75, 6eqtrd 2069 . . . . . 6 (𝑅 = 𝑆 → (𝑅𝑅) = (𝑆𝑆))
84, 7uneq12d 3092 . . . . 5 (𝑅 = 𝑆 → (𝑅 ∪ (𝑅𝑅)) = (𝑆 ∪ (𝑆𝑆)))
98sseq1d 2966 . . . 4 (𝑅 = 𝑆 → ((𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅 ↔ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑅))
10 sseq2 2961 . . . 4 (𝑅 = 𝑆 → ((𝑆 ∪ (𝑆𝑆)) ⊆ 𝑅 ↔ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆))
119, 10bitrd 177 . . 3 (𝑅 = 𝑆 → ((𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅 ↔ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆))
121, 3, 113anbi123d 1206 . 2 (𝑅 = 𝑆 → ((Rel 𝑅 dom 𝑅 = A (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅) ↔ (Rel 𝑆 dom 𝑆 = A (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆)))
13 df-er 6042 . 2 (𝑅 Er A ↔ (Rel 𝑅 dom 𝑅 = A (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅))
14 df-er 6042 . 2 (𝑆 Er A ↔ (Rel 𝑆 dom 𝑆 = A (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆))
1512, 13, 143bitr4g 212 1 (𝑅 = 𝑆 → (𝑅 Er A𝑆 Er A))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 98   w3a 884   = wceq 1242  cun 2909  wss 2911  ccnv 4287  dom cdm 4288  ccom 4292  Rel wrel 4293   Er wer 6039
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019
This theorem depends on definitions:  df-bi 110  df-3an 886  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-v 2553  df-un 2916  df-in 2918  df-ss 2925  df-sn 3373  df-pr 3374  df-op 3376  df-br 3756  df-opab 3810  df-rel 4295  df-cnv 4296  df-co 4297  df-dm 4298  df-er 6042
This theorem is referenced by:  riinerm  6115
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