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Theorem elssabg 4147
Description: Membership in a class abstraction involving a subset. Unlike elabg 2883, 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 4141 . . . 4 ((𝐴𝐵𝐵𝑉) → 𝐴 ∈ V)
21expcom 116 . . 3 (𝐵𝑉 → (𝐴𝐵𝐴 ∈ V))
32adantrd 279 . 2 (𝐵𝑉 → ((𝐴𝐵𝜓) → 𝐴 ∈ V))
4 sseq1 3178 . . . 4 (𝑥 = 𝐴 → (𝑥𝐵𝐴𝐵))
5 elssabg.1 . . . 4 (𝑥 = 𝐴 → (𝜑𝜓))
64, 5anbi12d 473 . . 3 (𝑥 = 𝐴 → ((𝑥𝐵𝜑) ↔ (𝐴𝐵𝜓)))
76elab3g 2888 . 2 (((𝐴𝐵𝜓) → 𝐴 ∈ V) → (𝐴 ∈ {𝑥 ∣ (𝑥𝐵𝜑)} ↔ (𝐴𝐵𝜓)))
83, 7syl 14 1 (𝐵𝑉 → (𝐴 ∈ {𝑥 ∣ (𝑥𝐵𝜑)} ↔ (𝐴𝐵𝜓)))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 104  wb 105   = wceq 1353  wcel 2148  {cab 2163  Vcvv 2737  wss 3129
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159  ax-sep 4120
This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-v 2739  df-in 3135  df-ss 3142
This theorem is referenced by: (None)
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