Theorem abssf 45155

Description: Class abstraction in a subclass relationship. (Contributed by Glauco Siliprandi, 26-Jun-2021.)

Hypothesis

Ref	Expression
abssf.1	⊢ Ⅎ𝑥𝐴

Assertion

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

Proof of Theorem abssf

Step	Hyp	Ref	Expression
1		abssf.1	. . . 4 ⊢ Ⅎ𝑥𝐴
2	1	abid2f 2925	. . 3 ⊢ {𝑥 ∣ 𝑥 ∈ 𝐴} = 𝐴
3	2	sseq2i 3964	. 2 ⊢ ({𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝑥 ∈ 𝐴} ↔ {𝑥 ∣ 𝜑} ⊆ 𝐴)
4		ss2ab 4013	. 2 ⊢ ({𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝑥 ∈ 𝐴} ↔ ∀𝑥(𝜑 → 𝑥 ∈ 𝐴))
5	3, 4	bitr3i 277	1 ⊢ ({𝑥 ∣ 𝜑} ⊆ 𝐴 ↔ ∀𝑥(𝜑 → 𝑥 ∈ 𝐴))

Syntax hints:   → wi 4   ↔ wb 206  ∀wal 1539   ∈ wcel 2111  {cab 2709  Ⅎwnfc 2879   ⊆ wss 3902

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-10 2144  ax-11 2160  ax-12 2180  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-ex 1781  df-nf 1785  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-nfc 2881  df-ss 3919

This theorem is referenced by:  rabssf  45162