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Theorem sbcbr2g 4086
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 4084 . 2 (𝐴𝐷 → ([𝐴 / 𝑥]𝐵𝑅𝐶𝐴 / 𝑥𝐵𝑅𝐴 / 𝑥𝐶))
2 csbconstg 3094 . . 3 (𝐴𝐷𝐴 / 𝑥𝐵 = 𝐵)
32breq1d 4039 . 2 (𝐴𝐷 → (𝐴 / 𝑥𝐵𝑅𝐴 / 𝑥𝐶𝐵𝑅𝐴 / 𝑥𝐶))
41, 3bitrd 188 1 (𝐴𝐷 → ([𝐴 / 𝑥]𝐵𝑅𝐶𝐵𝑅𝐴 / 𝑥𝐶))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 105  wcel 2164  [wsbc 2985  csb 3080   class class class wbr 4029
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-io 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2175
This theorem depends on definitions:  df-bi 117  df-3an 982  df-tru 1367  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-v 2762  df-sbc 2986  df-csb 3081  df-un 3157  df-sn 3624  df-pr 3625  df-op 3627  df-br 4030
This theorem is referenced by: (None)
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