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Theorem sbceq1g 4353
Description: Move proper substitution to first argument of an equality. (Contributed by NM, 30-Nov-2005.)
Assertion
Ref Expression
sbceq1g (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐶))
Distinct variable group:   𝑥,𝐶
Allowed substitution hints:   𝐴(𝑥)   𝐵(𝑥)   𝑉(𝑥)

Proof of Theorem sbceq1g
StepHypRef Expression
1 sbceqg 4348 . 2 (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶))
2 csbconstg 3855 . . 3 (𝐴𝑉𝐴 / 𝑥𝐶 = 𝐶)
32eqeq2d 2750 . 2 (𝐴𝑉 → (𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶𝐴 / 𝑥𝐵 = 𝐶))
41, 3bitrd 278 1 (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐶))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 205   = wceq 1541  wcel 2109  [wsbc 3719  csb 3836
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1801  ax-4 1815  ax-5 1916  ax-6 1974  ax-7 2014  ax-8 2111  ax-9 2119  ax-10 2140  ax-11 2157  ax-12 2174  ax-ext 2710
This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1544  df-ex 1786  df-nf 1790  df-sb 2071  df-clab 2717  df-cleq 2731  df-clel 2817  df-nfc 2890  df-v 3432  df-sbc 3720  df-csb 3837
This theorem is referenced by:  telgsums  19575  suppss2f  30953  f1od2  31035  finxpreclem4  35544  frege120  41544
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