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Theorem sbceq1g 4363
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 4358 . 2 (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶))
2 csbconstg 3899 . . 3 (𝐴𝑉𝐴 / 𝑥𝐶 = 𝐶)
32eqeq2d 2829 . 2 (𝐴𝑉 → (𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶𝐴 / 𝑥𝐵 = 𝐶))
41, 3bitrd 280 1 (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐶))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 207   = wceq 1528  wcel 2105  [wsbc 3769  csb 3880
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1787  ax-4 1801  ax-5 1902  ax-6 1961  ax-7 2006  ax-8 2107  ax-9 2115  ax-10 2136  ax-11 2151  ax-12 2167  ax-ext 2790
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 842  df-tru 1531  df-ex 1772  df-nf 1776  df-sb 2061  df-clab 2797  df-cleq 2811  df-clel 2890  df-nfc 2960  df-sbc 3770  df-csb 3881
This theorem is referenced by:  telgsums  19042  suppss2f  30312  f1od2  30383  finxpreclem4  34557  frege120  40207
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