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Theorem sbceq2g 4098
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 4092 . 2 (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶))
2 csbconstg 3652 . . 3 (𝐴𝑉𝐴 / 𝑥𝐵 = 𝐵)
32eqeq1d 2726 . 2 (𝐴𝑉 → (𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶𝐵 = 𝐴 / 𝑥𝐶))
41, 3bitrd 268 1 (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶𝐵 = 𝐴 / 𝑥𝐶))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 196   = wceq 1596  wcel 2103  [wsbc 3541  csb 3639
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1835  ax-4 1850  ax-5 1952  ax-6 2018  ax-7 2054  ax-9 2112  ax-10 2132  ax-11 2147  ax-12 2160  ax-13 2355  ax-ext 2704
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1599  df-ex 1818  df-nf 1823  df-sb 2011  df-clab 2711  df-cleq 2717  df-clel 2720  df-nfc 2855  df-v 3306  df-sbc 3542  df-csb 3640
This theorem is referenced by:  csbsng  4350  csbmpt12  5114  f1od2  29729  bj-snsetex  33178  csbmpt22g  33409  csbfinxpg  33457  poimirlem26  33667  cdlemkid3N  36640  cdlemkid4  36641  brtrclfv2  38438  frege116  38692
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