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Theorem sbcbr1g 3979
Description: Move substitution in and out of a binary relation. (Contributed by NM, 13-Dec-2005.)
Assertion
Ref Expression
sbcbr1g (𝐴𝐷 → ([𝐴 / 𝑥]𝐵𝑅𝐶𝐴 / 𝑥𝐵𝑅𝐶))
Distinct variable groups:   𝑥,𝐶   𝑥,𝑅
Allowed substitution hints:   𝐴(𝑥)   𝐵(𝑥)   𝐷(𝑥)

Proof of Theorem sbcbr1g
StepHypRef Expression
1 sbcbr12g 3978 . 2 (𝐴𝐷 → ([𝐴 / 𝑥]𝐵𝑅𝐶𝐴 / 𝑥𝐵𝑅𝐴 / 𝑥𝐶))
2 csbconstg 3011 . . 3 (𝐴𝐷𝐴 / 𝑥𝐶 = 𝐶)
32breq2d 3936 . 2 (𝐴𝐷 → (𝐴 / 𝑥𝐵𝑅𝐴 / 𝑥𝐶𝐴 / 𝑥𝐵𝑅𝐶))
41, 3bitrd 187 1 (𝐴𝐷 → ([𝐴 / 𝑥]𝐵𝑅𝐶𝐴 / 𝑥𝐵𝑅𝐶))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 104  wcel 1480  [wsbc 2904  csb 2998   class class class wbr 3924
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 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2119
This theorem depends on definitions:  df-bi 116  df-3an 964  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2124  df-cleq 2130  df-clel 2133  df-nfc 2268  df-v 2683  df-sbc 2905  df-csb 2999  df-un 3070  df-sn 3528  df-pr 3529  df-op 3531  df-br 3925
This theorem is referenced by: (None)
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