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Theorem ereq1 6542
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 4709 . . 3 (𝑅 = 𝑆 → (Rel 𝑅 ↔ Rel 𝑆))
2 dmeq 4828 . . . 4 (𝑅 = 𝑆 → dom 𝑅 = dom 𝑆)
32eqeq1d 2186 . . 3 (𝑅 = 𝑆 → (dom 𝑅 = 𝐴 ↔ dom 𝑆 = 𝐴))
4 cnveq 4802 . . . . . 6 (𝑅 = 𝑆𝑅 = 𝑆)
5 coeq1 4785 . . . . . . 7 (𝑅 = 𝑆 → (𝑅𝑅) = (𝑆𝑅))
6 coeq2 4786 . . . . . . 7 (𝑅 = 𝑆 → (𝑆𝑅) = (𝑆𝑆))
75, 6eqtrd 2210 . . . . . 6 (𝑅 = 𝑆 → (𝑅𝑅) = (𝑆𝑆))
84, 7uneq12d 3291 . . . . 5 (𝑅 = 𝑆 → (𝑅 ∪ (𝑅𝑅)) = (𝑆 ∪ (𝑆𝑆)))
98sseq1d 3185 . . . 4 (𝑅 = 𝑆 → ((𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅 ↔ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑅))
10 sseq2 3180 . . . 4 (𝑅 = 𝑆 → ((𝑆 ∪ (𝑆𝑆)) ⊆ 𝑅 ↔ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆))
119, 10bitrd 188 . . 3 (𝑅 = 𝑆 → ((𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅 ↔ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆))
121, 3, 113anbi123d 1312 . 2 (𝑅 = 𝑆 → ((Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅) ↔ (Rel 𝑆 ∧ dom 𝑆 = 𝐴 ∧ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆)))
13 df-er 6535 . 2 (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅))
14 df-er 6535 . 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 978   = wceq 1353  cun 3128  wss 3130  ccnv 4626  dom cdm 4627  ccom 4631  Rel wrel 4632   Er wer 6532
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 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159
This theorem depends on definitions:  df-bi 117  df-3an 980  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-v 2740  df-un 3134  df-in 3136  df-ss 3143  df-sn 3599  df-pr 3600  df-op 3602  df-br 4005  df-opab 4066  df-rel 4634  df-cnv 4635  df-co 4636  df-dm 4637  df-er 6535
This theorem is referenced by:  riinerm  6608
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