Theorem dfss7 4197

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 3917	. 2 ⊢ (𝐵 ⊆ 𝐴 ↔ (𝐵 ∩ 𝐴) = 𝐵)
2		dfin5 3907	. . . 4 ⊢ (𝐴 ∩ 𝐵) = {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵}
3	2	ineqcomi 4157	. . 3 ⊢ (𝐵 ∩ 𝐴) = {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵}
4	3	eqeq1i 2765	. 2 ⊢ ((𝐵 ∩ 𝐴) = 𝐵 ↔ {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵} = 𝐵)
5	1, 4	bitri 278	1 ⊢ (𝐵 ⊆ 𝐴 ↔ {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵} = 𝐵)

This proof depends on syntax axioms:   ↔ wb 209   = wceq 1570   ∈ wcel 2145  {crab 3412   ∩ cin 3898   ⊆ wss 3899

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-3an 1105  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-in 3906  df-ss 3916

This theorem is used by:  qusker  33789  nsgqusf1olem3  33844  f1oresf1orab  48177