Theorem brssr 38826

Description: The subset relation and subclass relationship (df-ss 3920) 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 38825	. . . . 5 ⊢ Rel S
2	1	brrelex1i 5688	. . . 4 ⊢ (𝐴 S 𝐵 → 𝐴 ∈ V)
3	2	adantl 481	. . 3 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐴 S 𝐵) → 𝐴 ∈ V)
4		simpl 482	. . 3 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐴 S 𝐵) → 𝐵 ∈ 𝑉)
5	3, 4	jca 511	. 2 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐴 S 𝐵) → (𝐴 ∈ V ∧ 𝐵 ∈ 𝑉))
6		ssexg 5270	. . . 4 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ 𝑉) → 𝐴 ∈ V)
7		simpr 484	. . . 4 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ 𝑉) → 𝐵 ∈ 𝑉)
8	6, 7	jca 511	. . 3 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ 𝑉) → (𝐴 ∈ V ∧ 𝐵 ∈ 𝑉))
9	8	ancoms 458	. 2 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐴 ⊆ 𝐵) → (𝐴 ∈ V ∧ 𝐵 ∈ 𝑉))
10		sseq1 3961	. . 3 ⊢ (𝑥 = 𝐴 → (𝑥 ⊆ 𝑦 ↔ 𝐴 ⊆ 𝑦))
11		sseq2 3962	. . 3 ⊢ (𝑦 = 𝐵 → (𝐴 ⊆ 𝑦 ↔ 𝐴 ⊆ 𝐵))
12		df-ssr 38823	. . 3 ⊢ S = {⟨𝑥, 𝑦⟩ ∣ 𝑥 ⊆ 𝑦}
13	10, 11, 12	brabg 5495	. 2 ⊢ ((𝐴 ∈ V ∧ 𝐵 ∈ 𝑉) → (𝐴 S 𝐵 ↔ 𝐴 ⊆ 𝐵))
14	5, 9, 13	pm5.21nd 802	1 ⊢ (𝐵 ∈ 𝑉 → (𝐴 S 𝐵 ↔ 𝐴 ⊆ 𝐵))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   ∈ wcel 2114  Vcvv 3442   ⊆ wss 3903   class class class wbr 5100   S cssr 38431

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709  ax-sep 5243  ax-pr 5379

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-ral 3053  df-rex 3063  df-rab 3402  df-v 3444  df-dif 3906  df-un 3908  df-in 3910  df-ss 3920  df-nul 4288  df-if 4482  df-sn 4583  df-pr 4585  df-op 4589  df-br 5101  df-opab 5163  df-xp 5638  df-rel 5639  df-ssr 38823

This theorem is referenced by:  brssrid  38827  brssrres  38829  brcnvssr  38831  extssr  38834  dfrefrels2  38838  dfsymrels2  38870  dftrrels2  38904