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Theorem releq 4365
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 2960 . 2 (A = B → (A ⊆ (V × V) ↔ B ⊆ (V × V)))
2 df-rel 4295 . 2 (Rel AA ⊆ (V × V))
3 df-rel 4295 . 2 (Rel BB ⊆ (V × V))
41, 2, 33bitr4g 212 1 (A = B → (Rel A ↔ Rel B))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 98   = wceq 1242  Vcvv 2551  wss 2911   × cxp 4286  Rel wrel 4293
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-11 1394  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019
This theorem depends on definitions:  df-bi 110  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-in 2918  df-ss 2925  df-rel 4295
This theorem is referenced by:  releqi  4366  releqd  4367  dfrel2  4714  tposfn2  5822  ereq1  6049
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