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Theorem elssabg 5349
Description: Membership in a class abstraction involving a subset. Unlike elabg 3677, 𝐴 does not have to be a set. (Contributed by NM, 29-Aug-2006.)
Hypothesis
Ref Expression
elssabg.1 (𝑥 = 𝐴 → (𝜑𝜓))
Assertion
Ref Expression
elssabg (𝐵𝑉 → (𝐴 ∈ {𝑥 ∣ (𝑥𝐵𝜑)} ↔ (𝐴𝐵𝜓)))
Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝜓,𝑥
Allowed substitution hints:   𝜑(𝑥)   𝑉(𝑥)

Proof of Theorem elssabg
StepHypRef Expression
1 ssexg 5329 . . . 4 ((𝐴𝐵𝐵𝑉) → 𝐴 ∈ V)
21expcom 413 . . 3 (𝐵𝑉 → (𝐴𝐵𝐴 ∈ V))
32adantrd 491 . 2 (𝐵𝑉 → ((𝐴𝐵𝜓) → 𝐴 ∈ V))
4 sseq1 4021 . . . 4 (𝑥 = 𝐴 → (𝑥𝐵𝐴𝐵))
5 elssabg.1 . . . 4 (𝑥 = 𝐴 → (𝜑𝜓))
64, 5anbi12d 632 . . 3 (𝑥 = 𝐴 → ((𝑥𝐵𝜑) ↔ (𝐴𝐵𝜓)))
76elab3g 3688 . 2 (((𝐴𝐵𝜓) → 𝐴 ∈ V) → (𝐴 ∈ {𝑥 ∣ (𝑥𝐵𝜑)} ↔ (𝐴𝐵𝜓)))
83, 7syl 17 1 (𝐵𝑉 → (𝐴 ∈ {𝑥 ∣ (𝑥𝐵𝜑)} ↔ (𝐴𝐵𝜓)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wa 395   = wceq 1537  wcel 2106  {cab 2712  Vcvv 3478  wss 3963
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706  ax-sep 5302
This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1088  df-tru 1540  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-rab 3434  df-v 3480  df-in 3970  df-ss 3980
This theorem is referenced by: (None)
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