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Theorem sbceq2g 4359
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 4352 . 2 (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶))
2 csbconstg 3856 . . 3 (𝐴𝑉𝐴 / 𝑥𝐵 = 𝐵)
32eqeq1d 2738 . 2 (𝐴𝑉 → (𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶𝐵 = 𝐴 / 𝑥𝐶))
41, 3bitrd 279 1 (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶𝐵 = 𝐴 / 𝑥𝐶))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206   = wceq 1542  wcel 2114  [wsbc 3728  csb 3837
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-10 2147  ax-11 2163  ax-12 2185  ax-ext 2708
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-ex 1782  df-nf 1786  df-sb 2069  df-clab 2715  df-cleq 2728  df-clel 2811  df-nfc 2885  df-v 3431  df-sbc 3729  df-csb 3838
This theorem is referenced by:  csbsng  4652  csbmpt12  5512  opsbc2ie  32545  f1od2  32792  bj-snsetex  37270  csbmpo123  37647  csbfinxpg  37704  poimirlem26  37967  cdlemkid3N  41379  cdlemkid4  41380  f1o2d2  42674  brtrclfv2  44154  frege116  44406
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