Theorem ssabf 44337

Description: Subclass of a class abstraction. (Contributed by Glauco Siliprandi, 26-Jun-2021.)

Hypothesis

Ref	Expression
ssabf.1	⊢ Ⅎ𝑥𝐴

Assertion

Ref	Expression
ssabf	⊢ (𝐴 ⊆ {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝜑))

Proof of Theorem ssabf

Step	Hyp	Ref	Expression
1		ssabf.1	. . . 4 ⊢ Ⅎ𝑥𝐴
2	1	abid2f 2928	. . 3 ⊢ {𝑥 ∣ 𝑥 ∈ 𝐴} = 𝐴
3	2	sseq1i 4003	. 2 ⊢ ({𝑥 ∣ 𝑥 ∈ 𝐴} ⊆ {𝑥 ∣ 𝜑} ↔ 𝐴 ⊆ {𝑥 ∣ 𝜑})
4		ss2ab 4049	. 2 ⊢ ({𝑥 ∣ 𝑥 ∈ 𝐴} ⊆ {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝜑))
5	3, 4	bitr3i 277	1 ⊢ (𝐴 ⊆ {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝜑))

Syntax hints:   → wi 4   ↔ wb 205  ∀wal 1531   ∈ wcel 2098  {cab 2701  Ⅎwnfc 2875   ⊆ wss 3941

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-10 2129  ax-11 2146  ax-12 2163  ax-ext 2695

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-tru 1536  df-ex 1774  df-nf 1778  df-sb 2060  df-clab 2702  df-cleq 2716  df-clel 2802  df-nfc 2877  df-v 3468  df-in 3948  df-ss 3958

This theorem is referenced by:  ssrabf  44351