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Theorem sbceq2g 4377
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 4370 . 2 (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶))
2 csbconstg 3866 . . 3 (𝐴𝑉𝐴 / 𝑥𝐵 = 𝐵)
32eqeq1d 2762 . 2 (𝐴𝑉 → (𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶𝐵 = 𝐴 / 𝑥𝐶))
41, 3bitrd 282 1 (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶𝐵 = 𝐴 / 𝑥𝐶))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  wi 4  wb 209   = wceq 1570  wcel 2145  [wsbc 3739  csb 3847
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-ext 2732
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-v 3452  df-sbc 3740  df-csb 3848
This theorem is used by:  csbsng  4669  csbmpt12  5536  mpof1o2d  8124  opsbc2ie  32952  f1od2  33191  bj-snsetex  37708  csbmpo123  38086  csbfinxpg  38143  poimirlem26  38396  cdlemkid3N  41807  cdlemkid4  41808  brtrclfv2  44568  frege116  44820
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