Theorem dfss7 4192

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 3908	. 2 ⊢ (𝐵 ⊆ 𝐴 ↔ (𝐵 ∩ 𝐴) = 𝐵)
2		dfin5 3898	. . . 4 ⊢ (𝐴 ∩ 𝐵) = {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵}
3	2	ineqcomi 4152	. . 3 ⊢ (𝐵 ∩ 𝐴) = {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵}
4	3	eqeq1i 2742	. 2 ⊢ ((𝐵 ∩ 𝐴) = 𝐵 ↔ {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵} = 𝐵)
5	1, 4	bitri 275	1 ⊢ (𝐵 ⊆ 𝐴 ↔ {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵} = 𝐵)

Syntax hints:   ↔ wb 206   = wceq 1542   ∈ wcel 2114  {crab 3390   ∩ cin 3889   ⊆ wss 3890

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1089  df-tru 1545  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-rab 3391  df-in 3897  df-ss 3907

This theorem is referenced by:  qusker  33424  nsgqusf1olem3  33490  f1oresf1orab  47749