Theorem dfss7 4180

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		df-ss 3909	. 2 ⊢ (𝐵 ⊆ 𝐴 ↔ (𝐵 ∩ 𝐴) = 𝐵)
2		incom 4140	. . . 4 ⊢ (𝐵 ∩ 𝐴) = (𝐴 ∩ 𝐵)
3		dfin5 3900	. . . 4 ⊢ (𝐴 ∩ 𝐵) = {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵}
4	2, 3	eqtri 2768	. . 3 ⊢ (𝐵 ∩ 𝐴) = {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵}
5	4	eqeq1i 2745	. 2 ⊢ ((𝐵 ∩ 𝐴) = 𝐵 ↔ {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵} = 𝐵)
6	1, 5	bitri 274	1 ⊢ (𝐵 ⊆ 𝐴 ↔ {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵} = 𝐵)

Syntax hints:   ↔ wb 205   = wceq 1542   ∈ wcel 2110  {crab 3070   ∩ cin 3891   ⊆ wss 3892

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 1975  ax-7 2015  ax-9 2120  ax-ext 2711

This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1545  df-ex 1787  df-sb 2072  df-clab 2718  df-cleq 2732  df-rab 3075  df-in 3899  df-ss 3909

This theorem is referenced by:  qusker  31545  nsgqusf1olem3  31596  f1oresf1orab  44749