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Theorem sbss 3329
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 2560 . 2 𝑦 ∈ V
2 sbequ 1721 . 2 (𝑧 = 𝑦 → ([𝑧 / 𝑥]𝑥𝐴 ↔ [𝑦 / 𝑥]𝑥𝐴))
3 sseq1 2966 . 2 (𝑧 = 𝑦 → (𝑧𝐴𝑦𝐴))
4 nfv 1421 . . 3 𝑥 𝑧𝐴
5 sseq1 2966 . . 3 (𝑥 = 𝑧 → (𝑥𝐴𝑧𝐴))
64, 5sbie 1674 . 2 ([𝑧 / 𝑥]𝑥𝐴𝑧𝐴)
71, 2, 3, 6vtoclb 2611 1 ([𝑦 / 𝑥]𝑥𝐴𝑦𝐴)
Colors of variables: wff set class
Syntax hints:  wb 98  [wsb 1645  wss 2917
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022
This theorem depends on definitions:  df-bi 110  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-v 2559  df-in 2924  df-ss 2931
This theorem is referenced by: (None)
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