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Theorem sbceq2g 2995
Description: Move proper substitution to second argument of an equality. (Contributed by NM, 30-Nov-2005.)
Assertion
Ref Expression
sbceq2g (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶𝐵 = 𝐴 / 𝑥𝐶))
Distinct variable group:   𝑥,𝐵
Allowed substitution hints:   𝐴(𝑥)   𝐶(𝑥)   𝑉(𝑥)

Proof of Theorem sbceq2g
StepHypRef Expression
1 sbceqg 2989 . 2 (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶))
2 csbconstg 2987 . . 3 (𝐴𝑉𝐴 / 𝑥𝐵 = 𝐵)
32eqeq1d 2126 . 2 (𝐴𝑉 → (𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶𝐵 = 𝐴 / 𝑥𝐶))
41, 3bitrd 187 1 (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶𝐵 = 𝐴 / 𝑥𝐶))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 104   = wceq 1316  wcel 1465  [wsbc 2882  csb 2975
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 683  ax-5 1408  ax-7 1409  ax-gen 1410  ax-ie1 1454  ax-ie2 1455  ax-8 1467  ax-10 1468  ax-11 1469  ax-i12 1470  ax-bndl 1471  ax-4 1472  ax-17 1491  ax-i9 1495  ax-ial 1499  ax-i5r 1500  ax-ext 2099
This theorem depends on definitions:  df-bi 116  df-tru 1319  df-nf 1422  df-sb 1721  df-clab 2104  df-cleq 2110  df-clel 2113  df-nfc 2247  df-v 2662  df-sbc 2883  df-csb 2976
This theorem is referenced by:  csbsng  3554  f1od2  6100
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