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Theorem sbss 3357
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 2605 . 2 𝑦 ∈ V
2 sbequ 1762 . 2 (𝑧 = 𝑦 → ([𝑧 / 𝑥]𝑥𝐴 ↔ [𝑦 / 𝑥]𝑥𝐴))
3 sseq1 3021 . 2 (𝑧 = 𝑦 → (𝑧𝐴𝑦𝐴))
4 nfv 1462 . . 3 𝑥 𝑧𝐴
5 sseq1 3021 . . 3 (𝑥 = 𝑧 → (𝑥𝐴𝑧𝐴))
64, 5sbie 1715 . 2 ([𝑧 / 𝑥]𝑥𝐴𝑧𝐴)
71, 2, 3, 6vtoclb 2657 1 ([𝑦 / 𝑥]𝑥𝐴𝑦𝐴)
Colors of variables: wff set class
Syntax hints:  wb 103  [wsb 1686  wss 2974
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2064
This theorem depends on definitions:  df-bi 115  df-nf 1391  df-sb 1687  df-clab 2069  df-cleq 2075  df-clel 2078  df-v 2604  df-in 2980  df-ss 2987
This theorem is referenced by: (None)
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