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Theorem sbss 3522
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 2733 . 2 𝑦 ∈ V
2 sbequ 1833 . 2 (𝑧 = 𝑦 → ([𝑧 / 𝑥]𝑥𝐴 ↔ [𝑦 / 𝑥]𝑥𝐴))
3 sseq1 3170 . 2 (𝑧 = 𝑦 → (𝑧𝐴𝑦𝐴))
4 nfv 1521 . . 3 𝑥 𝑧𝐴
5 sseq1 3170 . . 3 (𝑥 = 𝑧 → (𝑥𝐴𝑧𝐴))
64, 5sbie 1784 . 2 ([𝑧 / 𝑥]𝑥𝐴𝑧𝐴)
71, 2, 3, 6vtoclb 2787 1 ([𝑦 / 𝑥]𝑥𝐴𝑦𝐴)
Colors of variables: wff set class
Syntax hints:  wb 104  [wsb 1755  wss 3121
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152
This theorem depends on definitions:  df-bi 116  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-v 2732  df-in 3127  df-ss 3134
This theorem is referenced by: (None)
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