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Theorem releq 4628
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 3124 . 2 (𝐴 = 𝐵 → (𝐴 ⊆ (V × V) ↔ 𝐵 ⊆ (V × V)))
2 df-rel 4553 . 2 (Rel 𝐴𝐴 ⊆ (V × V))
3 df-rel 4553 . 2 (Rel 𝐵𝐵 ⊆ (V × V))
41, 2, 33bitr4g 222 1 (𝐴 = 𝐵 → (Rel 𝐴 ↔ Rel 𝐵))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 104   = wceq 1332  Vcvv 2689  wss 3075   × cxp 4544  Rel wrel 4551
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-11 1485  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516  ax-ext 2122
This theorem depends on definitions:  df-bi 116  df-nf 1438  df-sb 1737  df-clab 2127  df-cleq 2133  df-clel 2136  df-in 3081  df-ss 3088  df-rel 4553
This theorem is referenced by:  releqi  4629  releqd  4630  dfrel2  4996  tposfn2  6170  ereq1  6443
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