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Theorem sbss 4062
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 3193 . 2 𝑦 ∈ V
2 sbequ 2375 . 2 (𝑧 = 𝑦 → ([𝑧 / 𝑥]𝑥𝐴 ↔ [𝑦 / 𝑥]𝑥𝐴))
3 sseq1 3611 . 2 (𝑧 = 𝑦 → (𝑧𝐴𝑦𝐴))
4 nfv 1840 . . 3 𝑥 𝑧𝐴
5 sseq1 3611 . . 3 (𝑥 = 𝑧 → (𝑥𝐴𝑧𝐴))
64, 5sbie 2407 . 2 ([𝑧 / 𝑥]𝑥𝐴𝑧𝐴)
71, 2, 3, 6vtoclb 3253 1 ([𝑦 / 𝑥]𝑥𝐴𝑦𝐴)
Colors of variables: wff setvar class
Syntax hints:  wb 196  [wsb 1877  wss 3560
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1836  ax-6 1885  ax-7 1932  ax-9 1996  ax-10 2016  ax-11 2031  ax-12 2044  ax-13 2245  ax-ext 2601
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1483  df-ex 1702  df-nf 1707  df-sb 1878  df-clab 2608  df-cleq 2614  df-clel 2617  df-v 3192  df-in 3567  df-ss 3574
This theorem is referenced by: (None)
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