Theorem abssf 41371

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 3012	. . 3 ⊢ {𝑥 ∣ 𝑥 ∈ 𝐴} = 𝐴
3	2	sseq2i 3995	. 2 ⊢ ({𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝑥 ∈ 𝐴} ↔ {𝑥 ∣ 𝜑} ⊆ 𝐴)
4		ss2ab 4038	. 2 ⊢ ({𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝑥 ∈ 𝐴} ↔ ∀𝑥(𝜑 → 𝑥 ∈ 𝐴))
5	3, 4	bitr3i 279	1 ⊢ ({𝑥 ∣ 𝜑} ⊆ 𝐴 ↔ ∀𝑥(𝜑 → 𝑥 ∈ 𝐴))

Syntax hints:   → wi 4   ↔ wb 208  ∀wal 1531   ∈ wcel 2110  {cab 2799  Ⅎwnfc 2961   ⊆ wss 3935

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2157  ax-12 2173  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1536  df-ex 1777  df-nf 1781  df-sb 2066  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-in 3942  df-ss 3951

This theorem is referenced by:  rabssf  41378