Theorem dfss7 4270

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 3994	. 2 ⊢ (𝐵 ⊆ 𝐴 ↔ (𝐵 ∩ 𝐴) = 𝐵)
2		incom 4230	. . . 4 ⊢ (𝐵 ∩ 𝐴) = (𝐴 ∩ 𝐵)
3		dfin5 3984	. . . 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 275	1 ⊢ (𝐵 ⊆ 𝐴 ↔ {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵} = 𝐵)

Syntax hints:   ↔ wb 206   = wceq 1537   ∈ wcel 2108  {crab 3443   ∩ cin 3975   ⊆ wss 3976

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1089  df-tru 1540  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-rab 3444  df-in 3983  df-ss 3993

This theorem is referenced by:  qusker  33342  nsgqusf1olem3  33408  f1oresf1orab  47204