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Theorem bj-sbel1 33208
Description: Version of sbcel1g 4184 when substituting a set. (Note: one could have a corresponding version of sbcel12 4180 when substituting a set, but the point here is that the antecedent of sbcel1g 4184 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 3637 . 2 ([𝑦 / 𝑥]𝐴𝐵[𝑦 / 𝑥]𝐴𝐵)
2 vex 3394 . . 3 𝑦 ∈ V
3 sbcel1g 4184 . . 3 (𝑦 ∈ V → ([𝑦 / 𝑥]𝐴𝐵𝑦 / 𝑥𝐴𝐵))
42, 3ax-mp 5 . 2 ([𝑦 / 𝑥]𝐴𝐵𝑦 / 𝑥𝐴𝐵)
51, 4bitri 266 1 ([𝑦 / 𝑥]𝐴𝐵𝑦 / 𝑥𝐴𝐵)
Colors of variables: wff setvar class
Syntax hints:  wb 197  [wsb 2060  wcel 2156  Vcvv 3391  [wsbc 3633  csb 3728
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 2068  ax-7 2104  ax-9 2165  ax-10 2185  ax-11 2201  ax-12 2214  ax-13 2420  ax-ext 2784
This theorem depends on definitions:  df-bi 198  df-an 385  df-or 866  df-tru 1641  df-fal 1651  df-ex 1860  df-nf 1864  df-sb 2061  df-clab 2793  df-cleq 2799  df-clel 2802  df-nfc 2937  df-v 3393  df-sbc 3634  df-csb 3729  df-dif 3772  df-nul 4117
This theorem is referenced by:  bj-snsetex  33261
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