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Theorem releq 4765
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 3220 . 2 (𝐴 = 𝐵 → (𝐴 ⊆ (V × V) ↔ 𝐵 ⊆ (V × V)))
2 df-rel 4690 . 2 (Rel 𝐴𝐴 ⊆ (V × V))
3 df-rel 4690 . 2 (Rel 𝐵𝐵 ⊆ (V × V))
41, 2, 33bitr4g 223 1 (𝐴 = 𝐵 → (Rel 𝐴 ↔ Rel 𝐵))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 105   = wceq 1373  Vcvv 2773  wss 3170   × cxp 4681  Rel wrel 4688
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-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-11 1530  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-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-in 3176  df-ss 3183  df-rel 4690
This theorem is referenced by:  releqi  4766  releqd  4767  dfrel2  5142  tposfn2  6365  ereq1  6640
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