Theorem elssabg 4181

Description: Membership in a class abstraction involving a subset. Unlike elabg 2910, 𝐴 does not have to be a set. (Contributed by NM, 29-Aug-2006.)

Hypothesis

Ref	Expression
elssabg.1	⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
elssabg	⊢ (𝐵 ∈ 𝑉 → (𝐴 ∈ {𝑥 ∣ (𝑥 ⊆ 𝐵 ∧ 𝜑)} ↔ (𝐴 ⊆ 𝐵 ∧ 𝜓)))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝜓,𝑥

Allowed substitution hints:   𝜑(𝑥)   𝑉(𝑥)

Proof of Theorem elssabg

Step	Hyp	Ref	Expression
1		ssexg 4172	. . . 4 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ 𝑉) → 𝐴 ∈ V)
2	1	expcom 116	. . 3 ⊢ (𝐵 ∈ 𝑉 → (𝐴 ⊆ 𝐵 → 𝐴 ∈ V))
3	2	adantrd 279	. 2 ⊢ (𝐵 ∈ 𝑉 → ((𝐴 ⊆ 𝐵 ∧ 𝜓) → 𝐴 ∈ V))
4		sseq1 3206	. . . 4 ⊢ (𝑥 = 𝐴 → (𝑥 ⊆ 𝐵 ↔ 𝐴 ⊆ 𝐵))
5		elssabg.1	. . . 4 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
6	4, 5	anbi12d 473	. . 3 ⊢ (𝑥 = 𝐴 → ((𝑥 ⊆ 𝐵 ∧ 𝜑) ↔ (𝐴 ⊆ 𝐵 ∧ 𝜓)))
7	6	elab3g 2915	. 2 ⊢ (((𝐴 ⊆ 𝐵 ∧ 𝜓) → 𝐴 ∈ V) → (𝐴 ∈ {𝑥 ∣ (𝑥 ⊆ 𝐵 ∧ 𝜑)} ↔ (𝐴 ⊆ 𝐵 ∧ 𝜓)))
8	3, 7	syl 14	1 ⊢ (𝐵 ∈ 𝑉 → (𝐴 ∈ {𝑥 ∣ (𝑥 ⊆ 𝐵 ∧ 𝜑)} ↔ (𝐴 ⊆ 𝐵 ∧ 𝜓)))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105   = wceq 1364   ∈ wcel 2167  {cab 2182  Vcvv 2763   ⊆ wss 3157

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 710  ax-5 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-10 1519  ax-11 1520  ax-i12 1521  ax-bndl 1523  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549  ax-ext 2178  ax-sep 4151

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1475  df-sb 1777  df-clab 2183  df-cleq 2189  df-clel 2192  df-nfc 2328  df-v 2765  df-in 3163  df-ss 3170

This theorem is referenced by: (None)