Theorem brssr 39092

Description: The subset relation and subclass relationship (df-ss 3924) are the same, that is, (𝐴 S 𝐵 ↔ 𝐴 ⊆ 𝐵) when 𝐵 is a set. (Contributed by Peter Mazsa, 31-Jul-2019.)

Assertion

Ref	Expression
brssr	⊢ (𝐵 ∈ 𝑉 → (𝐴 S 𝐵 ↔ 𝐴 ⊆ 𝐵))

Proof of Theorem brssr

Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		relssr 39091	. . . . 5 ⊢ Rel S
2	1	brrelex1i 5708	. . . 4 ⊢ (𝐴 S 𝐵 → 𝐴 ∈ V)
3	2	adantl 486	. . 3 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐴 S 𝐵) → 𝐴 ∈ V)
4		simpl 487	. . 3 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐴 S 𝐵) → 𝐵 ∈ 𝑉)
5	3, 4	jca 520	. 2 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐴 S 𝐵) → (𝐴 ∈ V ∧ 𝐵 ∈ 𝑉))
6		ssexg 5284	. . . 4 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ 𝑉) → 𝐴 ∈ V)
7		simpr 489	. . . 4 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ 𝑉) → 𝐵 ∈ 𝑉)
8	6, 7	jca 520	. . 3 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ 𝑉) → (𝐴 ∈ V ∧ 𝐵 ∈ 𝑉))
9	8	ancoms 463	. 2 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐴 ⊆ 𝐵) → (𝐴 ∈ V ∧ 𝐵 ∈ 𝑉))
10		sseq1 3964	. . 3 ⊢ (𝑥 = 𝐴 → (𝑥 ⊆ 𝑦 ↔ 𝐴 ⊆ 𝑦))
11		sseq2 3965	. . 3 ⊢ (𝑦 = 𝐵 → (𝐴 ⊆ 𝑦 ↔ 𝐴 ⊆ 𝐵))
12		df-ssr 39089	. . 3 ⊢ S = {⟨𝑥, 𝑦⟩ ∣ 𝑥 ⊆ 𝑦}
13	10, 11, 12	brabg 5515	. 2 ⊢ ((𝐴 ∈ V ∧ 𝐵 ∈ 𝑉) → (𝐴 S 𝐵 ↔ 𝐴 ⊆ 𝐵))
14	5, 9, 13	pm5.21nd 813	1 ⊢ (𝐵 ∈ 𝑉 → (𝐴 S 𝐵 ↔ 𝐴 ⊆ 𝐵))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 400   ∈ wcel 2145  Vcvv 3457   ⊆ wss 3907   class class class wbr 5105   S cssr 38697

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031  ax-8 2147  ax-9 2155  ax-ext 2737  ax-sep 5251  ax-pr 5395

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1103  df-tru 1566  df-fal 1576  df-ex 1803  df-sb 2094  df-clab 2744  df-cleq 2757  df-clel 2840  df-ral 3080  df-rex 3090  df-rab 3418  df-v 3459  df-dif 3910  df-un 3912  df-in 3914  df-ss 3924  df-nul 4289  df-if 4484  df-sn 4586  df-pr 4588  df-op 4592  df-br 5106  df-opab 5168  df-xp 5658  df-rel 5659  df-ssr 39089

This theorem is referenced by:  brssrid  39093  brssrres  39095  brcnvssr  39097  extssr  39100  dfrefrels2  39104  dfsymrels2  39136  dftrrels2  39170