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Theorem sbceq1g 2965
 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 2961 . 2 (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶))
2 csbconstg 2959 . . 3 (𝐴𝑉𝐴 / 𝑥𝐶 = 𝐶)
32eqeq2d 2106 . 2 (𝐴𝑉 → (𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶𝐴 / 𝑥𝐵 = 𝐶))
41, 3bitrd 187 1 (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐶))
 Colors of variables: wff set class Syntax hints:   → wi 4   ↔ wb 104   = wceq 1296   ∈ wcel 1445  [wsbc 2854  ⦋csb 2947 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 668  ax-5 1388  ax-7 1389  ax-gen 1390  ax-ie1 1434  ax-ie2 1435  ax-8 1447  ax-10 1448  ax-11 1449  ax-i12 1450  ax-bndl 1451  ax-4 1452  ax-17 1471  ax-i9 1475  ax-ial 1479  ax-i5r 1480  ax-ext 2077 This theorem depends on definitions:  df-bi 116  df-tru 1299  df-nf 1402  df-sb 1700  df-clab 2082  df-cleq 2088  df-clel 2091  df-nfc 2224  df-v 2635  df-sbc 2855  df-csb 2948 This theorem is referenced by:  f1od2  6038
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