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Theorem sbcbr2g 4146
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 4144 . 2 (𝐴𝐷 → ([𝐴 / 𝑥]𝐵𝑅𝐶𝐴 / 𝑥𝐵𝑅𝐴 / 𝑥𝐶))
2 csbconstg 3141 . . 3 (𝐴𝐷𝐴 / 𝑥𝐵 = 𝐵)
32breq1d 4098 . 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 2202  [wsbc 3031  csb 3127   class class class wbr 4088
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 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-bndl 1557  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-ext 2213
This theorem depends on definitions:  df-bi 117  df-3an 1006  df-tru 1400  df-nf 1509  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2363  df-v 2804  df-sbc 3032  df-csb 3128  df-un 3204  df-sn 3675  df-pr 3676  df-op 3678  df-br 4089
This theorem is referenced by: (None)
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