Theorem sseqin2 3300

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 3285	1 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐵 ∩ 𝐴) = 𝐴)

Syntax hints:   ↔ wb 104   = wceq 1332   ∩ cin 3075   ⊆ wss 3076

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 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-bndl 1487  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516  ax-ext 2122

This theorem depends on definitions:  df-bi 116  df-tru 1335  df-nf 1438  df-sb 1737  df-clab 2127  df-cleq 2133  df-clel 2136  df-nfc 2271  df-v 2691  df-in 3082  df-ss 3089

This theorem is referenced by:  dfss4st  3314  resabs1  4856  rescnvcnv  5009  frecfnom  6306  fiintim  6825  nn0supp  9053  uzin  9382  iooval2  9728  fzval2  9824  dfphi2  11932  resttopon  12379  restabs  12383  restopnb  12389  txcnmpt  12481