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Theorem sbss 3659
 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]x Ay A)
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 2862 . 2 y V
2 sbequ 2060 . 2 (z = y → ([z / x]x A ↔ [y / x]x A))
3 sseq1 3292 . 2 (z = y → (z Ay A))
4 nfv 1619 . . 3 x z A
5 sseq1 3292 . . 3 (x = z → (x Az A))
64, 5sbie 2038 . 2 ([z / x]x Az A)
71, 2, 3, 6vtoclb 2912 1 ([y / x]x Ay A)
 Colors of variables: wff setvar class Syntax hints:   ↔ wb 176  [wsb 1648   ⊆ wss 3257 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334 This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2478  df-v 2861  df-nin 3211  df-compl 3212  df-in 3213  df-ss 3259 This theorem is referenced by: (None)
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