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Theorem elssabg 5297
Description: Membership in a class abstraction involving a subset. Unlike elabg 3632, 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 5284 . . . 4 ((𝐴𝐵𝐵𝑉) → 𝐴 ∈ V)
21expcom 415 . . 3 (𝐵𝑉 → (𝐴𝐵𝐴 ∈ V))
32adantrd 493 . 2 (𝐵𝑉 → ((𝐴𝐵𝜓) → 𝐴 ∈ V))
4 sseq1 3973 . . . 4 (𝑥 = 𝐴 → (𝑥𝐵𝐴𝐵))
5 elssabg.1 . . . 4 (𝑥 = 𝐴 → (𝜑𝜓))
64, 5anbi12d 632 . . 3 (𝑥 = 𝐴 → ((𝑥𝐵𝜑) ↔ (𝐴𝐵𝜓)))
76elab3g 3641 . 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 205  wa 397   = wceq 1542  wcel 2107  {cab 2710  Vcvv 3447  wss 3914
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2704  ax-sep 5260
This theorem depends on definitions:  df-bi 206  df-an 398  df-tru 1545  df-ex 1783  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-rab 3407  df-v 3449  df-in 3921  df-ss 3931
This theorem is referenced by: (None)
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