MPE Home Metamath Proof Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >  sbss Structured version   Visualization version   GIF version

Theorem sbss 4030
Description: Set substitution into the first argument of a subset relation. (Contributed by Rodolfo Medina, 7-Jul-2010.) (Proof shortened by Mario Carneiro, 14-Nov-2016.)
Assertion
Ref Expression
sbss ([𝑦 / 𝑥]𝑥𝐴𝑦𝐴)
Distinct variable group:   𝑥,𝐴
Allowed substitution hint:   𝐴(𝑦)

Proof of Theorem sbss
Dummy variable 𝑧 is distinct from all other variables.
StepHypRef Expression
1 vex 3172 . 2 𝑦 ∈ V
2 sbequ 2360 . 2 (𝑧 = 𝑦 → ([𝑧 / 𝑥]𝑥𝐴 ↔ [𝑦 / 𝑥]𝑥𝐴))
3 sseq1 3585 . 2 (𝑧 = 𝑦 → (𝑧𝐴𝑦𝐴))
4 nfv 1829 . . 3 𝑥 𝑧𝐴
5 sseq1 3585 . . 3 (𝑥 = 𝑧 → (𝑥𝐴𝑧𝐴))
64, 5sbie 2392 . 2 ([𝑧 / 𝑥]𝑥𝐴𝑧𝐴)
71, 2, 3, 6vtoclb 3232 1 ([𝑦 / 𝑥]𝑥𝐴𝑦𝐴)
Colors of variables: wff setvar class
Syntax hints:  wb 194  [wsb 1866  wss 3536
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1712  ax-4 1727  ax-5 1826  ax-6 1874  ax-7 1921  ax-10 2005  ax-11 2020  ax-12 2032  ax-13 2229  ax-ext 2586
This theorem depends on definitions:  df-bi 195  df-or 383  df-an 384  df-tru 1477  df-ex 1695  df-nf 1700  df-sb 1867  df-clab 2593  df-cleq 2599  df-clel 2602  df-v 3171  df-in 3543  df-ss 3550
This theorem is referenced by: (None)
  Copyright terms: Public domain W3C validator