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Theorem ereq1 6144
 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 4450 . . 3 (𝑅 = 𝑆 → (Rel 𝑅 ↔ Rel 𝑆))
2 dmeq 4563 . . . 4 (𝑅 = 𝑆 → dom 𝑅 = dom 𝑆)
32eqeq1d 2064 . . 3 (𝑅 = 𝑆 → (dom 𝑅 = 𝐴 ↔ dom 𝑆 = 𝐴))
4 cnveq 4537 . . . . . 6 (𝑅 = 𝑆𝑅 = 𝑆)
5 coeq1 4521 . . . . . . 7 (𝑅 = 𝑆 → (𝑅𝑅) = (𝑆𝑅))
6 coeq2 4522 . . . . . . 7 (𝑅 = 𝑆 → (𝑆𝑅) = (𝑆𝑆))
75, 6eqtrd 2088 . . . . . 6 (𝑅 = 𝑆 → (𝑅𝑅) = (𝑆𝑆))
84, 7uneq12d 3126 . . . . 5 (𝑅 = 𝑆 → (𝑅 ∪ (𝑅𝑅)) = (𝑆 ∪ (𝑆𝑆)))
98sseq1d 3000 . . . 4 (𝑅 = 𝑆 → ((𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅 ↔ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑅))
10 sseq2 2995 . . . 4 (𝑅 = 𝑆 → ((𝑆 ∪ (𝑆𝑆)) ⊆ 𝑅 ↔ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆))
119, 10bitrd 181 . . 3 (𝑅 = 𝑆 → ((𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅 ↔ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆))
121, 3, 113anbi123d 1218 . 2 (𝑅 = 𝑆 → ((Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅) ↔ (Rel 𝑆 ∧ dom 𝑆 = 𝐴 ∧ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆)))
13 df-er 6137 . 2 (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅))
14 df-er 6137 . 2 (𝑆 Er 𝐴 ↔ (Rel 𝑆 ∧ dom 𝑆 = 𝐴 ∧ (𝑆 ∪ (𝑆𝑆)) ⊆ 𝑆))
1512, 13, 143bitr4g 216 1 (𝑅 = 𝑆 → (𝑅 Er 𝐴𝑆 Er 𝐴))
 Colors of variables: wff set class Syntax hints:   → wi 4   ↔ wb 102   ∧ w3a 896   = wceq 1259   ∪ cun 2943   ⊆ wss 2945  ◡ccnv 4372  dom cdm 4373   ∘ ccom 4377  Rel wrel 4378   Er wer 6134 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-io 640  ax-5 1352  ax-7 1353  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-10 1412  ax-11 1413  ax-i12 1414  ax-bndl 1415  ax-4 1416  ax-17 1435  ax-i9 1439  ax-ial 1443  ax-i5r 1444  ax-ext 2038 This theorem depends on definitions:  df-bi 114  df-3an 898  df-tru 1262  df-nf 1366  df-sb 1662  df-clab 2043  df-cleq 2049  df-clel 2052  df-nfc 2183  df-v 2576  df-un 2950  df-in 2952  df-ss 2959  df-sn 3409  df-pr 3410  df-op 3412  df-br 3793  df-opab 3847  df-rel 4380  df-cnv 4381  df-co 4382  df-dm 4383  df-er 6137 This theorem is referenced by:  riinerm  6210
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