Users' Mathboxes Mathbox for BJ < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >   Mathboxes  >  bj-sbeqALT Structured version   Visualization version   GIF version

Theorem bj-sbeqALT 32870
Description: Substitution in an equality (use the more genereal version bj-sbeq 32871 instead, without disjoint variable condition). (Contributed by BJ, 6-Oct-2018.) (New usage is discouraged.) (Proof modification is discouraged.)
Assertion
Ref Expression
bj-sbeqALT ([𝑦 / 𝑥]𝐴 = 𝐵𝑦 / 𝑥𝐴 = 𝑦 / 𝑥𝐵)
Distinct variable group:   𝑥,𝑦
Allowed substitution hints:   𝐴(𝑥,𝑦)   𝐵(𝑥,𝑦)

Proof of Theorem bj-sbeqALT
StepHypRef Expression
1 nfcsb1v 3542 . . 3 𝑥𝑦 / 𝑥𝐴
2 nfcsb1v 3542 . . 3 𝑥𝑦 / 𝑥𝐵
31, 2nfeq 2773 . 2 𝑥𝑦 / 𝑥𝐴 = 𝑦 / 𝑥𝐵
4 csbeq1a 3535 . . 3 (𝑥 = 𝑦𝐴 = 𝑦 / 𝑥𝐴)
5 csbeq1a 3535 . . 3 (𝑥 = 𝑦𝐵 = 𝑦 / 𝑥𝐵)
64, 5eqeq12d 2635 . 2 (𝑥 = 𝑦 → (𝐴 = 𝐵𝑦 / 𝑥𝐴 = 𝑦 / 𝑥𝐵))
73, 6sbie 2406 1 ([𝑦 / 𝑥]𝐴 = 𝐵𝑦 / 𝑥𝐴 = 𝑦 / 𝑥𝐵)
Colors of variables: wff setvar class
Syntax hints:  wb 196   = wceq 1481  [wsb 1878  csb 3526
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1720  ax-4 1735  ax-5 1837  ax-6 1886  ax-7 1933  ax-9 1997  ax-10 2017  ax-11 2032  ax-12 2045  ax-13 2244  ax-ext 2600
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1484  df-ex 1703  df-nf 1708  df-sb 1879  df-clab 2607  df-cleq 2613  df-clel 2616  df-nfc 2751  df-sbc 3430  df-csb 3527
This theorem is referenced by: (None)
  Copyright terms: Public domain W3C validator