Theorem ssabf 45801

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 2955	. . 3 ⊢ {𝑥 ∣ 𝑥 ∈ 𝐴} = 𝐴
3	2	sseq1i 3966	. 2 ⊢ ({𝑥 ∣ 𝑥 ∈ 𝐴} ⊆ {𝑥 ∣ 𝜑} ↔ 𝐴 ⊆ {𝑥 ∣ 𝜑})
4		ss2ab 4016	. 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 1568   ∈ wcel 2143  {cab 2741  Ⅎwnfc 2910   ⊆ wss 3906

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-10 2176  ax-11 2192  ax-12 2213  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1573  df-ex 1810  df-nf 1814  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-nfc 2912  df-ss 3923

This theorem is referenced by:  ssrabf  45815