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Theorem releq 4686
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 3165 . 2 (𝐴 = 𝐵 → (𝐴 ⊆ (V × V) ↔ 𝐵 ⊆ (V × V)))
2 df-rel 4611 . 2 (Rel 𝐴𝐴 ⊆ (V × V))
3 df-rel 4611 . 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 1343  Vcvv 2726  wss 3116   × cxp 4602  Rel wrel 4609
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 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-11 1494  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147
This theorem depends on definitions:  df-bi 116  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-in 3122  df-ss 3129  df-rel 4611
This theorem is referenced by:  releqi  4687  releqd  4688  dfrel2  5054  tposfn2  6234  ereq1  6508
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