Theorem sseqin2 3425

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 3410	1 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐵 ∩ 𝐴) = 𝐴)

Syntax hints:   ↔ wb 105   = wceq 1397   ∩ cin 3198   ⊆ wss 3199

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 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-bndl 1557  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-ext 2212

This theorem depends on definitions:  df-bi 117  df-tru 1400  df-nf 1509  df-sb 1810  df-clab 2217  df-cleq 2223  df-clel 2226  df-nfc 2362  df-v 2803  df-in 3205  df-ss 3212

This theorem is referenced by:  dfss4st  3439  resabs1  5044  mptimass  5091  rescnvcnv  5201  frecfnom  6572  fiintim  7128  nn0supp  9459  uzin  9794  iooval2  10155  fzval2  10251  suprzubdc  10502  bitsinv1  12546  dfphi2  12815  ressabsg  13182  resttopon  14924  restabs  14928  restopnb  14934  txcnmpt  15026