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Theorem ereq1 6640
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 4765 . . 3 (𝑅 = 𝑆 → (Rel 𝑅 ↔ Rel 𝑆))
2 dmeq 4887 . . . 4 (𝑅 = 𝑆 → dom 𝑅 = dom 𝑆)
32eqeq1d 2215 . . 3 (𝑅 = 𝑆 → (dom 𝑅 = 𝐴 ↔ dom 𝑆 = 𝐴))
4 cnveq 4860 . . . . . 6 (𝑅 = 𝑆𝑅 = 𝑆)
5 coeq1 4843 . . . . . . 7 (𝑅 = 𝑆 → (𝑅𝑅) = (𝑆𝑅))
6 coeq2 4844 . . . . . . 7 (𝑅 = 𝑆 → (𝑆𝑅) = (𝑆𝑆))
75, 6eqtrd 2239 . . . . . 6 (𝑅 = 𝑆 → (𝑅𝑅) = (𝑆𝑆))
84, 7uneq12d 3332 . . . . 5 (𝑅 = 𝑆 → (𝑅 ∪ (𝑅𝑅)) = (𝑆 ∪ (𝑆𝑆)))
98sseq1d 3226 . . . 4 (𝑅 = 𝑆 → ((𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅 ↔ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑅))
10 sseq2 3221 . . . 4 (𝑅 = 𝑆 → ((𝑆 ∪ (𝑆𝑆)) ⊆ 𝑅 ↔ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆))
119, 10bitrd 188 . . 3 (𝑅 = 𝑆 → ((𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅 ↔ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆))
121, 3, 113anbi123d 1325 . 2 (𝑅 = 𝑆 → ((Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅) ↔ (Rel 𝑆 ∧ dom 𝑆 = 𝐴 ∧ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆)))
13 df-er 6633 . 2 (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅))
14 df-er 6633 . 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 981   = wceq 1373  cun 3168  wss 3170  ccnv 4682  dom cdm 4683  ccom 4687  Rel wrel 4688   Er wer 6630
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 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2188
This theorem depends on definitions:  df-bi 117  df-3an 983  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-nfc 2338  df-v 2775  df-un 3174  df-in 3176  df-ss 3183  df-sn 3644  df-pr 3645  df-op 3647  df-br 4052  df-opab 4114  df-rel 4690  df-cnv 4691  df-co 4692  df-dm 4693  df-er 6633
This theorem is referenced by:  riinerm  6708
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