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Theorem ereq1 6752
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 4814 . . 3 (𝑅 = 𝑆 → (Rel 𝑅 ↔ Rel 𝑆))
2 dmeq 4937 . . . 4 (𝑅 = 𝑆 → dom 𝑅 = dom 𝑆)
32eqeq1d 2240 . . 3 (𝑅 = 𝑆 → (dom 𝑅 = 𝐴 ↔ dom 𝑆 = 𝐴))
4 cnveq 4910 . . . . . 6 (𝑅 = 𝑆𝑅 = 𝑆)
5 coeq1 4893 . . . . . . 7 (𝑅 = 𝑆 → (𝑅𝑅) = (𝑆𝑅))
6 coeq2 4894 . . . . . . 7 (𝑅 = 𝑆 → (𝑆𝑅) = (𝑆𝑆))
75, 6eqtrd 2264 . . . . . 6 (𝑅 = 𝑆 → (𝑅𝑅) = (𝑆𝑆))
84, 7uneq12d 3364 . . . . 5 (𝑅 = 𝑆 → (𝑅 ∪ (𝑅𝑅)) = (𝑆 ∪ (𝑆𝑆)))
98sseq1d 3257 . . . 4 (𝑅 = 𝑆 → ((𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅 ↔ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑅))
10 sseq2 3252 . . . 4 (𝑅 = 𝑆 → ((𝑆 ∪ (𝑆𝑆)) ⊆ 𝑅 ↔ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆))
119, 10bitrd 188 . . 3 (𝑅 = 𝑆 → ((𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅 ↔ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆))
121, 3, 113anbi123d 1349 . 2 (𝑅 = 𝑆 → ((Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅) ↔ (Rel 𝑆 ∧ dom 𝑆 = 𝐴 ∧ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆)))
13 df-er 6745 . 2 (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅))
14 df-er 6745 . 2 (𝑆 Er 𝐴 ↔ (Rel 𝑆 ∧ dom 𝑆 = 𝐴 ∧ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆))
1512, 13, 143bitr4g 223 1 (𝑅 = 𝑆 → (𝑅 Er 𝐴𝑆 Er 𝐴))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 105  w3a 1005   = wceq 1398  cun 3199  wss 3201  ccnv 4730  dom cdm 4731  ccom 4735  Rel wrel 4736   Er wer 6742
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2213
This theorem depends on definitions:  df-bi 117  df-3an 1007  df-tru 1401  df-nf 1510  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2364  df-v 2805  df-un 3205  df-in 3207  df-ss 3214  df-sn 3679  df-pr 3680  df-op 3682  df-br 4094  df-opab 4156  df-rel 4738  df-cnv 4739  df-co 4740  df-dm 4741  df-er 6745
This theorem is referenced by:  riinerm  6820
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