Theorem ssabf 41736

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 2984	. . 3 ⊢ {𝑥 ∣ 𝑥 ∈ 𝐴} = 𝐴
3	2	sseq1i 3943	. 2 ⊢ ({𝑥 ∣ 𝑥 ∈ 𝐴} ⊆ {𝑥 ∣ 𝜑} ↔ 𝐴 ⊆ {𝑥 ∣ 𝜑})
4		ss2ab 3987	. 2 ⊢ ({𝑥 ∣ 𝑥 ∈ 𝐴} ⊆ {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝜑))
5	3, 4	bitr3i 280	1 ⊢ (𝐴 ⊆ {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝜑))

Syntax hints:   → wi 4   ↔ wb 209  ∀wal 1536   ∈ wcel 2111  {cab 2776  Ⅎwnfc 2936   ⊆ wss 3881

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 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-10 2142  ax-11 2158  ax-12 2175  ax-ext 2770

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-tru 1541  df-ex 1782  df-nf 1786  df-sb 2070  df-clab 2777  df-cleq 2791  df-clel 2870  df-nfc 2938  df-v 3443  df-in 3888  df-ss 3898

This theorem is referenced by:  ssrabf  41750