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Theorem elssabg 5225
Description: Membership in a class abstraction involving a subset. Unlike elabg 3652, 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 5213 . . . 4 ((𝐴𝐵𝐵𝑉) → 𝐴 ∈ V)
21expcom 417 . . 3 (𝐵𝑉 → (𝐴𝐵𝐴 ∈ V))
32adantrd 495 . 2 (𝐵𝑉 → ((𝐴𝐵𝜓) → 𝐴 ∈ V))
4 sseq1 3978 . . . 4 (𝑥 = 𝐴 → (𝑥𝐵𝐴𝐵))
5 elssabg.1 . . . 4 (𝑥 = 𝐴 → (𝜑𝜓))
64, 5anbi12d 633 . . 3 (𝑥 = 𝐴 → ((𝑥𝐵𝜑) ↔ (𝐴𝐵𝜓)))
76elab3g 3659 . 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 209  wa 399   = wceq 1538  wcel 2115  {cab 2802  Vcvv 3480  wss 3919
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1971  ax-7 2016  ax-8 2117  ax-9 2125  ax-10 2146  ax-11 2162  ax-12 2179  ax-ext 2796  ax-sep 5189
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-tru 1541  df-ex 1782  df-nf 1786  df-sb 2071  df-clab 2803  df-cleq 2817  df-clel 2896  df-nfc 2964  df-rab 3142  df-v 3482  df-in 3926  df-ss 3936
This theorem is referenced by: (None)
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