Theorem elssabg 4134

Description: Membership in a class abstraction involving a subset. Unlike elabg 2876, 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 4128	. . . 4 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ 𝑉) → 𝐴 ∈ V)
2	1	expcom 115	. . 3 ⊢ (𝐵 ∈ 𝑉 → (𝐴 ⊆ 𝐵 → 𝐴 ∈ V))
3	2	adantrd 277	. 2 ⊢ (𝐵 ∈ 𝑉 → ((𝐴 ⊆ 𝐵 ∧ 𝜓) → 𝐴 ∈ V))
4		sseq1 3170	. . . 4 ⊢ (𝑥 = 𝐴 → (𝑥 ⊆ 𝐵 ↔ 𝐴 ⊆ 𝐵))
5		elssabg.1	. . . 4 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
6	4, 5	anbi12d 470	. . 3 ⊢ (𝑥 = 𝐴 → ((𝑥 ⊆ 𝐵 ∧ 𝜑) ↔ (𝐴 ⊆ 𝐵 ∧ 𝜓)))
7	6	elab3g 2881	. 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 103   ↔ wb 104   = wceq 1348   ∈ wcel 2141  {cab 2156  Vcvv 2730   ⊆ wss 3121

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152  ax-sep 4107

This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-v 2732  df-in 3127  df-ss 3134

This theorem is referenced by: (None)