Theorem abssf 46221

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 2952	. . 3 ⊢ {𝑥 ∣ 𝑥 ∈ 𝐴} = 𝐴
3	2	sseq2i 3937	. 2 ⊢ ({𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝑥 ∈ 𝐴} ↔ {𝑥 ∣ 𝜑} ⊆ 𝐴)
4		ss2ab 3986	. 2 ⊢ ({𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝑥 ∈ 𝐴} ↔ ∀𝑥(𝜑 → 𝑥 ∈ 𝐴))
5	3, 4	bitr3i 280	1 ⊢ ({𝑥 ∣ 𝜑} ⊆ 𝐴 ↔ ∀𝑥(𝜑 → 𝑥 ∈ 𝐴))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1567   ∈ wcel 2144  {cab 2738  Ⅎwnfc 2907   ⊆ wss 3895

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-11 2193  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-ss 3896

This theorem is used by:  rabssf  46228