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Theorem releq 4508
Description: Equality theorem for the relation predicate. (Contributed by NM, 1-Aug-1994.)
Assertion
Ref Expression
releq (𝐴 = 𝐵 → (Rel 𝐴 ↔ Rel 𝐵))

Proof of Theorem releq
StepHypRef Expression
1 sseq1 3045 . 2 (𝐴 = 𝐵 → (𝐴 ⊆ (V × V) ↔ 𝐵 ⊆ (V × V)))
2 df-rel 4435 . 2 (Rel 𝐴𝐴 ⊆ (V × V))
3 df-rel 4435 . 2 (Rel 𝐵𝐵 ⊆ (V × V))
41, 2, 33bitr4g 221 1 (𝐴 = 𝐵 → (Rel 𝐴 ↔ Rel 𝐵))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 103   = wceq 1289  Vcvv 2619  wss 2997   × cxp 4426  Rel wrel 4433
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1381  ax-7 1382  ax-gen 1383  ax-ie1 1427  ax-ie2 1428  ax-8 1440  ax-11 1442  ax-4 1445  ax-17 1464  ax-i9 1468  ax-ial 1472  ax-i5r 1473  ax-ext 2070
This theorem depends on definitions:  df-bi 115  df-nf 1395  df-sb 1693  df-clab 2075  df-cleq 2081  df-clel 2084  df-in 3003  df-ss 3010  df-rel 4435
This theorem is referenced by:  releqi  4509  releqd  4510  dfrel2  4868  tposfn2  6013  ereq1  6279
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