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Theorem sbcbr2g 4105
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 4103 . 2 (𝐴𝐷 → ([𝐴 / 𝑥]𝐵𝑅𝐶𝐴 / 𝑥𝐵𝑅𝐴 / 𝑥𝐶))
2 csbconstg 3108 . . 3 (𝐴𝐷𝐴 / 𝑥𝐵 = 𝐵)
32breq1d 4057 . 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 2177  [wsbc 2999  csb 3094   class class class wbr 4047
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 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2188
This theorem depends on definitions:  df-bi 117  df-3an 983  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-nfc 2338  df-v 2775  df-sbc 3000  df-csb 3095  df-un 3171  df-sn 3640  df-pr 3641  df-op 3643  df-br 4048
This theorem is referenced by: (None)
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