Theorem dfss7

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

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

Proof

Step	Hyp	Ref	Expression
1		dfss2	⊢ ( 𝐵 ⊆ 𝐴 ↔ ( 𝐵 ∩ 𝐴 ) = 𝐵 )
2		incom	⊢ ( 𝐵 ∩ 𝐴 ) = ( 𝐴 ∩ 𝐵 )
3		dfin5	⊢ ( 𝐴 ∩ 𝐵 ) = { 𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵 }
4	2 3	eqtri	⊢ ( 𝐵 ∩ 𝐴 ) = { 𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵 }
5	4	eqeq1i	⊢ ( ( 𝐵 ∩ 𝐴 ) = 𝐵 ↔ { 𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵 } = 𝐵 )
6	1 5	bitri	⊢ ( 𝐵 ⊆ 𝐴 ↔ { 𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵 } = 𝐵 )