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Theorem elssabg 3893
Description: Membership in a class abstraction involving a subset. Unlike elabg 2682, 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 (B 𝑉 → (A {x ∣ (xB φ)} ↔ (AB ψ)))
Distinct variable groups:   x,A   x,B   ψ,x
Allowed substitution hints:   φ(x)   𝑉(x)

Proof of Theorem elssabg
StepHypRef Expression
1 ssexg 3887 . . . 4 ((AB B 𝑉) → A V)
21expcom 109 . . 3 (B 𝑉 → (ABA V))
32adantrd 264 . 2 (B 𝑉 → ((AB ψ) → A V))
4 sseq1 2960 . . . 4 (x = A → (xBAB))
5 elssabg.1 . . . 4 (x = A → (φψ))
64, 5anbi12d 442 . . 3 (x = A → ((xB φ) ↔ (AB ψ)))
76elab3g 2687 . 2 (((AB ψ) → A V) → (A {x ∣ (xB φ)} ↔ (AB ψ)))
83, 7syl 14 1 (B 𝑉 → (A {x ∣ (xB φ)} ↔ (AB ψ)))
Colors of variables: wff set class
Syntax hints:  wi 4   wa 97  wb 98   = wceq 1242   wcel 1390  {cab 2023  Vcvv 2551  wss 2911
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019  ax-sep 3866
This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-v 2553  df-in 2918  df-ss 2925
This theorem is referenced by: (None)
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