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Theorem 1st2ndbr 6163
Description: Express an element of a relation as a relationship between first and second components. (Contributed by Mario Carneiro, 22-Jun-2016.)
Assertion
Ref Expression
1st2ndbr ((Rel 𝐵𝐴𝐵) → (1st𝐴)𝐵(2nd𝐴))

Proof of Theorem 1st2ndbr
StepHypRef Expression
1 1st2nd 6160 . . 3 ((Rel 𝐵𝐴𝐵) → 𝐴 = ⟨(1st𝐴), (2nd𝐴)⟩)
2 simpr 109 . . 3 ((Rel 𝐵𝐴𝐵) → 𝐴𝐵)
31, 2eqeltrrd 2248 . 2 ((Rel 𝐵𝐴𝐵) → ⟨(1st𝐴), (2nd𝐴)⟩ ∈ 𝐵)
4 df-br 3990 . 2 ((1st𝐴)𝐵(2nd𝐴) ↔ ⟨(1st𝐴), (2nd𝐴)⟩ ∈ 𝐵)
53, 4sylibr 133 1 ((Rel 𝐵𝐴𝐵) → (1st𝐴)𝐵(2nd𝐴))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 103  wcel 2141  cop 3586   class class class wbr 3989  Rel wrel 4616  cfv 5198  1st c1st 6117  2nd c2nd 6118
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-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-13 2143  ax-14 2144  ax-ext 2152  ax-sep 4107  ax-pow 4160  ax-pr 4194  ax-un 4418
This theorem depends on definitions:  df-bi 116  df-3an 975  df-tru 1351  df-nf 1454  df-sb 1756  df-eu 2022  df-mo 2023  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-ral 2453  df-rex 2454  df-v 2732  df-sbc 2956  df-un 3125  df-in 3127  df-ss 3134  df-pw 3568  df-sn 3589  df-pr 3590  df-op 3592  df-uni 3797  df-br 3990  df-opab 4051  df-mpt 4052  df-id 4278  df-xp 4617  df-rel 4618  df-cnv 4619  df-co 4620  df-dm 4621  df-rn 4622  df-iota 5160  df-fun 5200  df-fv 5206  df-1st 6119  df-2nd 6120
This theorem is referenced by: (None)
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