Theorem ssab 3274

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 2330	. . 3 ⊢ {𝑥 ∣ 𝑥 ∈ 𝐴} = 𝐴
2	1	sseq1i 3230	. 2 ⊢ ({𝑥 ∣ 𝑥 ∈ 𝐴} ⊆ {𝑥 ∣ 𝜑} ↔ 𝐴 ⊆ {𝑥 ∣ 𝜑})
3		ss2ab 3272	. 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 1373   ∈ wcel 2180  {cab 2195   ⊆ wss 3177

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 713  ax-5 1473  ax-7 1474  ax-gen 1475  ax-ie1 1519  ax-ie2 1520  ax-8 1530  ax-10 1531  ax-11 1532  ax-i12 1533  ax-bndl 1535  ax-4 1536  ax-17 1552  ax-i9 1556  ax-ial 1560  ax-i5r 1561  ax-ext 2191

This theorem depends on definitions:  df-bi 117  df-nf 1487  df-sb 1789  df-clab 2196  df-cleq 2202  df-clel 2205  df-nfc 2341  df-in 3183  df-ss 3190

This theorem is referenced by:  ssabral  3275  ssrab  3282