Theorem dfss7 4226

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 3944	. 2 ⊢ (𝐵 ⊆ 𝐴 ↔ (𝐵 ∩ 𝐴) = 𝐵)
2		incom 4184	. . . 4 ⊢ (𝐵 ∩ 𝐴) = (𝐴 ∩ 𝐵)
3		dfin5 3934	. . . 4 ⊢ (𝐴 ∩ 𝐵) = {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵}
4	2, 3	eqtri 2758	. . 3 ⊢ (𝐵 ∩ 𝐴) = {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵}
5	4	eqeq1i 2740	. 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 1540   ∈ wcel 2108  {crab 3415   ∩ cin 3925   ⊆ wss 3926

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2707

This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1088  df-tru 1543  df-ex 1780  df-sb 2065  df-clab 2714  df-cleq 2727  df-clel 2809  df-rab 3416  df-in 3933  df-ss 3943

This theorem is referenced by:  qusker  33364  nsgqusf1olem3  33430  f1oresf1orab  47318