Theorem elssabg 5303

Description: Membership in a class abstraction involving a subset. Unlike elabg 3629, 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 5280	. . . 4 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ 𝑉) → 𝐴 ∈ V)
2	1	expcom 419	. . 3 ⊢ (𝐵 ∈ 𝑉 → (𝐴 ⊆ 𝐵 → 𝐴 ∈ V))
3	2	adantrd 497	. 2 ⊢ (𝐵 ∈ 𝑉 → ((𝐴 ⊆ 𝐵 ∧ 𝜓) → 𝐴 ∈ V))
4		sseq1 3933	. . . 4 ⊢ (𝑥 = 𝐴 → (𝑥 ⊆ 𝐵 ↔ 𝐴 ⊆ 𝐵))
5		elssabg.1	. . . 4 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
6	4, 5	anbi12d 644	. . 3 ⊢ (𝑥 = 𝐴 → ((𝑥 ⊆ 𝐵 ∧ 𝜑) ↔ (𝐴 ⊆ 𝐵 ∧ 𝜓)))
7	6	elab3g 3638	. 2 ⊢ (((𝐴 ⊆ 𝐵 ∧ 𝜓) → 𝐴 ∈ V) → (𝐴 ∈ {𝑥 ∣ (𝑥 ⊆ 𝐵 ∧ 𝜑)} ↔ (𝐴 ⊆ 𝐵 ∧ 𝜓)))
8	3, 7	syl 18	1 ⊢ (𝐵 ∈ 𝑉 → (𝐴 ∈ {𝑥 ∣ (𝑥 ⊆ 𝐵 ∧ 𝜑)} ↔ (𝐴 ⊆ 𝐵 ∧ 𝜓)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1569   ∈ wcel 2144  {cab 2738  Vcvv 3450   ⊆ wss 3895

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732  ax-sep 5248

This proof depends on definitions:  df-bi 210  df-an 402  df-3an 1105  df-tru 1572  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452  df-ss 3896  df-in 4091

This theorem is used by: (None)