Theorem dfss7 4196

Description: Alternate definition of subclass relationship. (Contributed by AV, 1-Aug-2022.)

Assertion

Ref	Expression
dfss7	⊢ (𝐵 ⊆ 𝐴 ↔ {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵} = 𝐵)

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Proof of Theorem dfss7

Step	Hyp	Ref	Expression
1		dfss2 3915	. 2 ⊢ (𝐵 ⊆ 𝐴 ↔ (𝐵 ∩ 𝐴) = 𝐵)
2		incom 4154	. . . 4 ⊢ (𝐵 ∩ 𝐴) = (𝐴 ∩ 𝐵)
3		dfin5 3905	. . . 4 ⊢ (𝐴 ∩ 𝐵) = {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵}
4	2, 3	eqtri 2754	. . 3 ⊢ (𝐵 ∩ 𝐴) = {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵}
5	4	eqeq1i 2736	. 2 ⊢ ((𝐵 ∩ 𝐴) = 𝐵 ↔ {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵} = 𝐵)
6	1, 5	bitri 275	1 ⊢ (𝐵 ⊆ 𝐴 ↔ {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵} = 𝐵)

Syntax hints:   ↔ wb 206   = wceq 1541   ∈ wcel 2111  {crab 3395   ∩ cin 3896   ⊆ wss 3897

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 2113  ax-9 2121  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1088  df-tru 1544  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-rab 3396  df-in 3904  df-ss 3914

This theorem is referenced by:  qusker  33306  nsgqusf1olem3  33372  f1oresf1orab  47320