Theorem brssr 38955

Description: The subset relation and subclass relationship (df-ss 3907) 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 38954	. . . . 5 ⊢ Rel S
2	1	brrelex1i 5681	. . . 4 ⊢ (𝐴 S 𝐵 → 𝐴 ∈ V)
3	2	adantl 482	. . 3 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐴 S 𝐵) → 𝐴 ∈ V)
4		simpl 483	. . 3 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐴 S 𝐵) → 𝐵 ∈ 𝑉)
5	3, 4	jca 516	. 2 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐴 S 𝐵) → (𝐴 ∈ V ∧ 𝐵 ∈ 𝑉))
6		ssexg 5258	. . . 4 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ 𝑉) → 𝐴 ∈ V)
7		simpr 485	. . . 4 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ 𝑉) → 𝐵 ∈ 𝑉)
8	6, 7	jca 516	. . 3 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ 𝑉) → (𝐴 ∈ V ∧ 𝐵 ∈ 𝑉))
9	8	ancoms 459	. 2 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐴 ⊆ 𝐵) → (𝐴 ∈ V ∧ 𝐵 ∈ 𝑉))
10		sseq1 3947	. . 3 ⊢ (𝑥 = 𝐴 → (𝑥 ⊆ 𝑦 ↔ 𝐴 ⊆ 𝑦))
11		sseq2 3948	. . 3 ⊢ (𝑦 = 𝐵 → (𝐴 ⊆ 𝑦 ↔ 𝐴 ⊆ 𝐵))
12		df-ssr 38952	. . 3 ⊢ S = {⟨𝑥, 𝑦⟩ ∣ 𝑥 ⊆ 𝑦}
13	10, 11, 12	brabg 5488	. 2 ⊢ ((𝐴 ∈ V ∧ 𝐵 ∈ 𝑉) → (𝐴 S 𝐵 ↔ 𝐴 ⊆ 𝐵))
14	5, 9, 13	pm5.21nd 807	1 ⊢ (𝐵 ∈ 𝑉 → (𝐴 S 𝐵 ↔ 𝐴 ⊆ 𝐵))

Syntax hints:   → wi 4   ↔ wb 207   ∧ wa 396   ∈ wcel 2119  Vcvv 3432   ⊆ wss 3890   class class class wbr 5079   S cssr 38560

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2712  ax-sep 5225  ax-pr 5369

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-3an 1094  df-tru 1550  df-fal 1560  df-ex 1787  df-sb 2074  df-clab 2719  df-cleq 2732  df-clel 2815  df-ral 3055  df-rex 3065  df-rab 3393  df-v 3434  df-dif 3893  df-un 3895  df-in 3897  df-ss 3907  df-nul 4269  df-if 4462  df-sn 4563  df-pr 4565  df-op 4569  df-br 5080  df-opab 5142  df-xp 5631  df-rel 5632  df-ssr 38952

This theorem is referenced by:  brssrid  38956  brssrres  38958  brcnvssr  38960  extssr  38963  dfrefrels2  38967  dfsymrels2  38999  dftrrels2  39033