Theorem ssab 3249

Description: Subclass of a class abstraction. (Contributed by NM, 16-Aug-2006.)

Assertion

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

Distinct variable group:   𝑥,𝐴

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem ssab

Step	Hyp	Ref	Expression
1		abid2 2314	. . 3 ⊢ {𝑥 ∣ 𝑥 ∈ 𝐴} = 𝐴
2	1	sseq1i 3205	. 2 ⊢ ({𝑥 ∣ 𝑥 ∈ 𝐴} ⊆ {𝑥 ∣ 𝜑} ↔ 𝐴 ⊆ {𝑥 ∣ 𝜑})
3		ss2ab 3247	. 2 ⊢ ({𝑥 ∣ 𝑥 ∈ 𝐴} ⊆ {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝜑))
4	2, 3	bitr3i 186	1 ⊢ (𝐴 ⊆ {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝜑))

Syntax hints:   → wi 4   ↔ wb 105  ∀wal 1362   ∈ wcel 2164  {cab 2179   ⊆ wss 3153

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2175

This theorem depends on definitions:  df-bi 117  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-in 3159  df-ss 3166

This theorem is referenced by:  ssabral  3250  ssrab  3257