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Theorem sbceq2g 4371
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 4364 . 2 (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶))
2 csbconstg 3869 . . 3 (𝐴𝑉𝐴 / 𝑥𝐵 = 𝐵)
32eqeq1d 2739 . 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 205   = wceq 1541  wcel 2106  [wsbc 3734  csb 3850
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 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-10 2137  ax-11 2154  ax-12 2171  ax-ext 2708
This theorem depends on definitions:  df-bi 206  df-an 398  df-or 846  df-tru 1544  df-ex 1782  df-nf 1786  df-sb 2068  df-clab 2715  df-cleq 2729  df-clel 2815  df-nfc 2887  df-v 3445  df-sbc 3735  df-csb 3851
This theorem is referenced by:  csbsng  4664  csbmpt12  5508  opsbc2ie  31177  f1od2  31407  bj-snsetex  35290  csbmpo123  35658  csbfinxpg  35715  poimirlem26  35959  cdlemkid3N  39252  cdlemkid4  39253  brtrclfv2  41708  frege116  41960
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