Theorem sseqin2 3340

Description: A relationship between subclass and intersection. Similar to Exercise 9 of [TakeutiZaring] p. 18. (Contributed by NM, 17-May-1994.)

Assertion

Ref	Expression
sseqin2	⊢ (𝐴 ⊆ 𝐵 ↔ (𝐵 ∩ 𝐴) = 𝐴)

Proof of Theorem sseqin2

Step	Hyp	Ref	Expression
1		dfss1 3325	1 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐵 ∩ 𝐴) = 𝐴)

Syntax hints:   ↔ wb 104   = wceq 1343   ∩ cin 3114   ⊆ wss 3115

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 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2296  df-v 2727  df-in 3121  df-ss 3128

This theorem is referenced by:  dfss4st  3354  resabs1  4912  rescnvcnv  5065  frecfnom  6365  fiintim  6890  nn0supp  9162  uzin  9494  iooval2  9847  fzval2  9943  suprzubdc  11881  dfphi2  12148  resttopon  12771  restabs  12775  restopnb  12781  txcnmpt  12873