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Theorem ereq1 6012
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 4337 . . 3 (𝑅 = 𝑆 → (Rel 𝑅 ↔ Rel 𝑆))
2 dmeq 4450 . . . 4 (𝑅 = 𝑆 → dom 𝑅 = dom 𝑆)
32eqeq1d 2021 . . 3 (𝑅 = 𝑆 → (dom 𝑅 = A ↔ dom 𝑆 = A))
4 cnveq 4424 . . . . . 6 (𝑅 = 𝑆𝑅 = 𝑆)
5 coeq1 4408 . . . . . . 7 (𝑅 = 𝑆 → (𝑅𝑅) = (𝑆𝑅))
6 coeq2 4409 . . . . . . 7 (𝑅 = 𝑆 → (𝑆𝑅) = (𝑆𝑆))
75, 6eqtrd 2045 . . . . . 6 (𝑅 = 𝑆 → (𝑅𝑅) = (𝑆𝑆))
84, 7uneq12d 3066 . . . . 5 (𝑅 = 𝑆 → (𝑅 ∪ (𝑅𝑅)) = (𝑆 ∪ (𝑆𝑆)))
98sseq1d 2940 . . . 4 (𝑅 = 𝑆 → ((𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅 ↔ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑅))
10 sseq2 2935 . . . 4 (𝑅 = 𝑆 → ((𝑆 ∪ (𝑆𝑆)) ⊆ 𝑅 ↔ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆))
119, 10bitrd 177 . . 3 (𝑅 = 𝑆 → ((𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅 ↔ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆))
121, 3, 113anbi123d 1187 . 2 (𝑅 = 𝑆 → ((Rel 𝑅 dom 𝑅 = A (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅) ↔ (Rel 𝑆 dom 𝑆 = A (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆)))
13 df-er 6005 . 2 (𝑅 Er A ↔ (Rel 𝑅 dom 𝑅 = A (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅))
14 df-er 6005 . 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 867   = wceq 1223  cun 2883  wss 2885  ccnv 4259  dom cdm 4260  ccom 4264  Rel wrel 4265   Er wer 6002
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 614  ax-5 1309  ax-7 1310  ax-gen 1311  ax-ie1 1355  ax-ie2 1356  ax-8 1368  ax-10 1369  ax-11 1370  ax-i12 1371  ax-bnd 1372  ax-4 1373  ax-17 1392  ax-i9 1396  ax-ial 1400  ax-i5r 1401  ax-ext 1995
This theorem depends on definitions:  df-bi 110  df-3an 869  df-tru 1226  df-nf 1323  df-sb 1619  df-clab 2000  df-cleq 2006  df-clel 2009  df-nfc 2140  df-v 2528  df-un 2890  df-in 2892  df-ss 2899  df-sn 3345  df-pr 3346  df-op 3348  df-br 3728  df-opab 3782  df-rel 4267  df-cnv 4268  df-co 4269  df-dm 4270  df-er 6005
This theorem is referenced by:  riinerm  6078
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