Theorem elssabg 5260

Description: Membership in a class abstraction involving a subset. Unlike elabg 3607, 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 5247	. . . 4 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ 𝑉) → 𝐴 ∈ V)
2	1	expcom 414	. . 3 ⊢ (𝐵 ∈ 𝑉 → (𝐴 ⊆ 𝐵 → 𝐴 ∈ V))
3	2	adantrd 492	. 2 ⊢ (𝐵 ∈ 𝑉 → ((𝐴 ⊆ 𝐵 ∧ 𝜓) → 𝐴 ∈ V))
4		sseq1 3946	. . . 4 ⊢ (𝑥 = 𝐴 → (𝑥 ⊆ 𝐵 ↔ 𝐴 ⊆ 𝐵))
5		elssabg.1	. . . 4 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
6	4, 5	anbi12d 631	. . 3 ⊢ (𝑥 = 𝐴 → ((𝑥 ⊆ 𝐵 ∧ 𝜑) ↔ (𝐴 ⊆ 𝐵 ∧ 𝜓)))
7	6	elab3g 3616	. 2 ⊢ (((𝐴 ⊆ 𝐵 ∧ 𝜓) → 𝐴 ∈ V) → (𝐴 ∈ {𝑥 ∣ (𝑥 ⊆ 𝐵 ∧ 𝜑)} ↔ (𝐴 ⊆ 𝐵 ∧ 𝜓)))
8	3, 7	syl 17	1 ⊢ (𝐵 ∈ 𝑉 → (𝐴 ∈ {𝑥 ∣ (𝑥 ⊆ 𝐵 ∧ 𝜑)} ↔ (𝐴 ⊆ 𝐵 ∧ 𝜓)))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 396   = wceq 1539   ∈ wcel 2106  {cab 2715  Vcvv 3432   ⊆ wss 3887

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 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709  ax-sep 5223

This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1542  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-rab 3073  df-v 3434  df-in 3894  df-ss 3904

This theorem is referenced by: (None)