Theorem ssabral 3199

Description: The relation for a subclass of a class abstraction is equivalent to restricted quantification. (Contributed by NM, 6-Sep-2006.)

Assertion

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

Distinct variable group:   𝑥,𝐴

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem ssabral

Step	Hyp	Ref	Expression
1		ssab 3198	. 2 ⊢ (𝐴 ⊆ {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝜑))
2		df-ral 2440	. 2 ⊢ (∀𝑥 ∈ 𝐴 𝜑 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝜑))
3	1, 2	bitr4i 186	1 ⊢ (𝐴 ⊆ {𝑥 ∣ 𝜑} ↔ ∀𝑥 ∈ 𝐴 𝜑)

Syntax hints:   → wi 4   ↔ wb 104  ∀wal 1333   ∈ wcel 2128  {cab 2143  ∀wral 2435   ⊆ wss 3102

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1427  ax-7 1428  ax-gen 1429  ax-ie1 1473  ax-ie2 1474  ax-8 1484  ax-10 1485  ax-11 1486  ax-i12 1487  ax-bndl 1489  ax-4 1490  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2139

This theorem depends on definitions:  df-bi 116  df-nf 1441  df-sb 1743  df-clab 2144  df-cleq 2150  df-clel 2153  df-nfc 2288  df-ral 2440  df-in 3108  df-ss 3115

This theorem is referenced by:  txdis1cn  12649  bj-bdfindis  13493