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Theorem ereq1 6484
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 4667 . . 3 (𝑅 = 𝑆 → (Rel 𝑅 ↔ Rel 𝑆))
2 dmeq 4785 . . . 4 (𝑅 = 𝑆 → dom 𝑅 = dom 𝑆)
32eqeq1d 2166 . . 3 (𝑅 = 𝑆 → (dom 𝑅 = 𝐴 ↔ dom 𝑆 = 𝐴))
4 cnveq 4759 . . . . . 6 (𝑅 = 𝑆𝑅 = 𝑆)
5 coeq1 4742 . . . . . . 7 (𝑅 = 𝑆 → (𝑅𝑅) = (𝑆𝑅))
6 coeq2 4743 . . . . . . 7 (𝑅 = 𝑆 → (𝑆𝑅) = (𝑆𝑆))
75, 6eqtrd 2190 . . . . . 6 (𝑅 = 𝑆 → (𝑅𝑅) = (𝑆𝑆))
84, 7uneq12d 3262 . . . . 5 (𝑅 = 𝑆 → (𝑅 ∪ (𝑅𝑅)) = (𝑆 ∪ (𝑆𝑆)))
98sseq1d 3157 . . . 4 (𝑅 = 𝑆 → ((𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅 ↔ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑅))
10 sseq2 3152 . . . 4 (𝑅 = 𝑆 → ((𝑆 ∪ (𝑆𝑆)) ⊆ 𝑅 ↔ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆))
119, 10bitrd 187 . . 3 (𝑅 = 𝑆 → ((𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅 ↔ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆))
121, 3, 113anbi123d 1294 . 2 (𝑅 = 𝑆 → ((Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅) ↔ (Rel 𝑆 ∧ dom 𝑆 = 𝐴 ∧ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆)))
13 df-er 6477 . 2 (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅))
14 df-er 6477 . 2 (𝑆 Er 𝐴 ↔ (Rel 𝑆 ∧ dom 𝑆 = 𝐴 ∧ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆))
1512, 13, 143bitr4g 222 1 (𝑅 = 𝑆 → (𝑅 Er 𝐴𝑆 Er 𝐴))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 104  w3a 963   = wceq 1335  cun 3100  wss 3102  ccnv 4584  dom cdm 4585  ccom 4589  Rel wrel 4590   Er wer 6474
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1427  ax-7 1428  ax-gen 1429  ax-ie1 1473  ax-ie2 1474  ax-8 1484  ax-10 1485  ax-11 1486  ax-i12 1487  ax-bndl 1489  ax-4 1490  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2139
This theorem depends on definitions:  df-bi 116  df-3an 965  df-tru 1338  df-nf 1441  df-sb 1743  df-clab 2144  df-cleq 2150  df-clel 2153  df-nfc 2288  df-v 2714  df-un 3106  df-in 3108  df-ss 3115  df-sn 3566  df-pr 3567  df-op 3569  df-br 3966  df-opab 4026  df-rel 4592  df-cnv 4593  df-co 4594  df-dm 4595  df-er 6477
This theorem is referenced by:  riinerm  6550
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