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Theorem sbceq1g 4325
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 4320 . 2 (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶))
2 csbconstg 3850 . . 3 (𝐴𝑉𝐴 / 𝑥𝐶 = 𝐶)
32eqeq2d 2812 . 2 (𝐴𝑉 → (𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶𝐴 / 𝑥𝐵 = 𝐶))
41, 3bitrd 282 1 (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐶))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 209   = wceq 1538  wcel 2112  [wsbc 3723  csb 3831
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 1911  ax-6 1970  ax-7 2015  ax-8 2114  ax-9 2122  ax-10 2143  ax-11 2159  ax-12 2176  ax-ext 2773
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-tru 1541  df-ex 1782  df-nf 1786  df-sb 2070  df-clab 2780  df-cleq 2794  df-clel 2873  df-nfc 2941  df-sbc 3724  df-csb 3832
This theorem is referenced by:  telgsums  19109  suppss2f  30401  f1od2  30486  finxpreclem4  34806  frege120  40671
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