Theorem brssr 36546

Description: The subset relation and subclass relationship (df-ss 3900) 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 36545	. . . . 5 ⊢ Rel S
2	1	brrelex1i 5634	. . . 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 5242	. . . 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 3942	. . 3 ⊢ (𝑥 = 𝐴 → (𝑥 ⊆ 𝑦 ↔ 𝐴 ⊆ 𝑦))
11		sseq2 3943	. . 3 ⊢ (𝑦 = 𝐵 → (𝐴 ⊆ 𝑦 ↔ 𝐴 ⊆ 𝐵))
12		df-ssr 36543	. . 3 ⊢ S = {⟨𝑥, 𝑦⟩ ∣ 𝑥 ⊆ 𝑦}
13	10, 11, 12	brabg 5445	. 2 ⊢ ((𝐴 ∈ V ∧ 𝐵 ∈ 𝑉) → (𝐴 S 𝐵 ↔ 𝐴 ⊆ 𝐵))
14	5, 9, 13	pm5.21nd 798	1 ⊢ (𝐵 ∈ 𝑉 → (𝐴 S 𝐵 ↔ 𝐴 ⊆ 𝐵))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395   ∈ wcel 2108  Vcvv 3422   ⊆ wss 3883   class class class wbr 5070   S cssr 36263

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709  ax-sep 5218  ax-nul 5225  ax-pr 5347

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-3an 1087  df-tru 1542  df-fal 1552  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-ral 3068  df-rex 3069  df-rab 3072  df-v 3424  df-dif 3886  df-un 3888  df-in 3890  df-ss 3900  df-nul 4254  df-if 4457  df-sn 4559  df-pr 4561  df-op 4565  df-br 5071  df-opab 5133  df-xp 5586  df-rel 5587  df-ssr 36543

This theorem is referenced by:  brssrid  36547  brssrres  36549  brcnvssr  36551  extssr  36554  dfrefrels2  36558  dfsymrels2  36586  dftrrels2  36616