Theorem brssr 38613

Description: The subset relation and subclass relationship (df-ss 3915) 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 38612	. . . . 5 ⊢ Rel S
2	1	brrelex1i 5675	. . . 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 5263	. . . 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 3956	. . 3 ⊢ (𝑥 = 𝐴 → (𝑥 ⊆ 𝑦 ↔ 𝐴 ⊆ 𝑦))
11		sseq2 3957	. . 3 ⊢ (𝑦 = 𝐵 → (𝐴 ⊆ 𝑦 ↔ 𝐴 ⊆ 𝐵))
12		df-ssr 38610	. . 3 ⊢ S = {⟨𝑥, 𝑦⟩ ∣ 𝑥 ⊆ 𝑦}
13	10, 11, 12	brabg 5482	. 2 ⊢ ((𝐴 ∈ V ∧ 𝐵 ∈ 𝑉) → (𝐴 S 𝐵 ↔ 𝐴 ⊆ 𝐵))
14	5, 9, 13	pm5.21nd 801	1 ⊢ (𝐵 ∈ 𝑉 → (𝐴 S 𝐵 ↔ 𝐴 ⊆ 𝐵))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   ∈ wcel 2113  Vcvv 3437   ⊆ wss 3898   class class class wbr 5093   S cssr 38245

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-ext 2705  ax-sep 5236  ax-nul 5246  ax-pr 5372

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2712  df-cleq 2725  df-clel 2808  df-ral 3049  df-rex 3058  df-rab 3397  df-v 3439  df-dif 3901  df-un 3903  df-in 3905  df-ss 3915  df-nul 4283  df-if 4475  df-sn 4576  df-pr 4578  df-op 4582  df-br 5094  df-opab 5156  df-xp 5625  df-rel 5626  df-ssr 38610

This theorem is referenced by:  brssrid  38614  brssrres  38616  brcnvssr  38618  extssr  38621  dfrefrels2  38625  dfsymrels2  38657  dftrrels2  38691