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Theorem sbcbr2g 3947
Description: Move substitution in and out of a binary relation. (Contributed by NM, 13-Dec-2005.)
Assertion
Ref Expression
sbcbr2g (𝐴𝐷 → ([𝐴 / 𝑥]𝐵𝑅𝐶𝐵𝑅𝐴 / 𝑥𝐶))
Distinct variable groups:   𝑥,𝐵   𝑥,𝑅
Allowed substitution hints:   𝐴(𝑥)   𝐶(𝑥)   𝐷(𝑥)

Proof of Theorem sbcbr2g
StepHypRef Expression
1 sbcbr12g 3945 . 2 (𝐴𝐷 → ([𝐴 / 𝑥]𝐵𝑅𝐶𝐴 / 𝑥𝐵𝑅𝐴 / 𝑥𝐶))
2 csbconstg 2983 . . 3 (𝐴𝐷𝐴 / 𝑥𝐵 = 𝐵)
32breq1d 3905 . 2 (𝐴𝐷 → (𝐴 / 𝑥𝐵𝑅𝐴 / 𝑥𝐶𝐵𝑅𝐴 / 𝑥𝐶))
41, 3bitrd 187 1 (𝐴𝐷 → ([𝐴 / 𝑥]𝐵𝑅𝐶𝐵𝑅𝐴 / 𝑥𝐶))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 104  wcel 1463  [wsbc 2878  csb 2971   class class class wbr 3895
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 681  ax-5 1406  ax-7 1407  ax-gen 1408  ax-ie1 1452  ax-ie2 1453  ax-8 1465  ax-10 1466  ax-11 1467  ax-i12 1468  ax-bndl 1469  ax-4 1470  ax-17 1489  ax-i9 1493  ax-ial 1497  ax-i5r 1498  ax-ext 2097
This theorem depends on definitions:  df-bi 116  df-3an 947  df-tru 1317  df-nf 1420  df-sb 1719  df-clab 2102  df-cleq 2108  df-clel 2111  df-nfc 2244  df-v 2659  df-sbc 2879  df-csb 2972  df-un 3041  df-sn 3499  df-pr 3500  df-op 3502  df-br 3896
This theorem is referenced by: (None)
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