Theorem ssexgOLD 5299

Description: Obsolete version of ssexg 5295 as of 18-Jul-2026. (Contributed by NM, 14-Aug-1994.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
ssexgOLD	⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ 𝐶) → 𝐴 ∈ V)

Proof of Theorem ssexgOLD

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		sseq2 3966	. . . 4 ⊢ (𝑥 = 𝐵 → (𝐴 ⊆ 𝑥 ↔ 𝐴 ⊆ 𝐵))
2	1	imbi1d 344	. . 3 ⊢ (𝑥 = 𝐵 → ((𝐴 ⊆ 𝑥 → 𝐴 ∈ V) ↔ (𝐴 ⊆ 𝐵 → 𝐴 ∈ V)))
3		vex 3462	. . . 4 ⊢ 𝑥 ∈ V
4	3	ssex 5296	. . 3 ⊢ (𝐴 ⊆ 𝑥 → 𝐴 ∈ V)
5	2, 4	vtoclg 3525	. 2 ⊢ (𝐵 ∈ 𝐶 → (𝐴 ⊆ 𝐵 → 𝐴 ∈ V))
6	5	impcom 413	1 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ 𝐶) → 𝐴 ∈ V)

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1570   ∈ wcel 2146  Vcvv 3458   ⊆ wss 3908

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2738  ax-sep 5262

This proof depends on definitions:  df-bi 210  df-an 402  df-3an 1105  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-rab 3420  df-v 3460  df-in 3915  df-ss 3925

This theorem is used by: (None)