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Theorem sbceq1g 3156
 Description: Move proper substitution to first argument of an equality. (Contributed by NM, 30-Nov-2005.)
Assertion
Ref Expression
sbceq1g (A V → ([̣A / xB = C[A / x]B = C))
Distinct variable group:   x,C
Allowed substitution hints:   A(x)   B(x)   V(x)

Proof of Theorem sbceq1g
StepHypRef Expression
1 sbceqg 3152 . 2 (A V → ([̣A / xB = C[A / x]B = [A / x]C))
2 csbconstg 3150 . . 3 (A V[A / x]C = C)
32eqeq2d 2364 . 2 (A V → ([A / x]B = [A / x]C[A / x]B = C))
41, 3bitrd 244 1 (A V → ([̣A / xB = C[A / x]B = C))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 176   = wceq 1642   ∈ wcel 1710  [̣wsbc 3046  [csb 3136 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334 This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2478  df-v 2861  df-sbc 3047  df-csb 3137 This theorem is referenced by:  eqerlem  5960
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