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Theorem sbceq1g 4196
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 4192 . 2 (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶))
2 csbconstg 3752 . . 3 (𝐴𝑉𝐴 / 𝑥𝐶 = 𝐶)
32eqeq2d 2827 . 2 (𝐴𝑉 → (𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶𝐴 / 𝑥𝐵 = 𝐶))
41, 3bitrd 270 1 (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐶))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 197   = wceq 1637  wcel 2157  [wsbc 3644  csb 3739
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1877  ax-4 1894  ax-5 2001  ax-6 2069  ax-7 2105  ax-9 2166  ax-10 2186  ax-11 2202  ax-12 2215  ax-13 2422  ax-ext 2795
This theorem depends on definitions:  df-bi 198  df-an 385  df-or 866  df-tru 1641  df-ex 1860  df-nf 1864  df-sb 2062  df-clab 2804  df-cleq 2810  df-clel 2813  df-nfc 2948  df-v 3404  df-sbc 3645  df-csb 3740
This theorem is referenced by:  telgsums  18599  suppss2f  29776  f1od2  29836  finxpreclem4  33553  frege120  38782
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