Theorem ssabf 41386

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 3012	. . 3 ⊢ {𝑥 ∣ 𝑥 ∈ 𝐴} = 𝐴
3	2	sseq1i 3995	. 2 ⊢ ({𝑥 ∣ 𝑥 ∈ 𝐴} ⊆ {𝑥 ∣ 𝜑} ↔ 𝐴 ⊆ {𝑥 ∣ 𝜑})
4		ss2ab 4039	. 2 ⊢ ({𝑥 ∣ 𝑥 ∈ 𝐴} ⊆ {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝜑))
5	3, 4	bitr3i 279	1 ⊢ (𝐴 ⊆ {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝜑))

Syntax hints:   → wi 4   ↔ wb 208  ∀wal 1535   ∈ wcel 2114  {cab 2799  Ⅎwnfc 2961   ⊆ wss 3936

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 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2161  ax-12 2177  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1540  df-ex 1781  df-nf 1785  df-sb 2070  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-in 3943  df-ss 3952

This theorem is referenced by:  ssrabf  41401