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Theorem bj-sbel1 32092
Description: Version of sbcel1g 3939 when substituting a set. (Note: one could have a corresponding version of sbcel12 3935 when substituting a set, but the point here is that the antecedent of sbcel1g 3939 is not needed when substituting a set.) (Contributed by BJ, 6-Oct-2018.)
Assertion
Ref Expression
bj-sbel1 ([𝑦 / 𝑥]𝐴𝐵𝑦 / 𝑥𝐴𝐵)
Distinct variable group:   𝑥,𝐵
Allowed substitution hints:   𝐴(𝑥,𝑦)   𝐵(𝑦)

Proof of Theorem bj-sbel1
StepHypRef Expression
1 sbsbc 3406 . 2 ([𝑦 / 𝑥]𝐴𝐵[𝑦 / 𝑥]𝐴𝐵)
2 vex 3176 . . 3 𝑦 ∈ V
3 sbcel1g 3939 . . 3 (𝑦 ∈ V → ([𝑦 / 𝑥]𝐴𝐵𝑦 / 𝑥𝐴𝐵))
42, 3ax-mp 5 . 2 ([𝑦 / 𝑥]𝐴𝐵𝑦 / 𝑥𝐴𝐵)
51, 4bitri 263 1 ([𝑦 / 𝑥]𝐴𝐵𝑦 / 𝑥𝐴𝐵)
Colors of variables: wff setvar class
Syntax hints:  wb 195  [wsb 1867  wcel 1977  Vcvv 3173  [wsbc 3402  csb 3499
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590
This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-fal 1481  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-v 3175  df-sbc 3403  df-csb 3500  df-dif 3543  df-nul 3875
This theorem is referenced by:  bj-snsetex  32144
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