Theorem sseqin2 3424

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 3409	1 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐵 ∩ 𝐴) = 𝐴)

Syntax hints:   ↔ wb 105   = wceq 1395   ∩ cin 3197   ⊆ wss 3198

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 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-v 2802  df-in 3204  df-ss 3211

This theorem is referenced by:  dfss4st  3438  resabs1  5040  mptimass  5087  rescnvcnv  5197  frecfnom  6562  fiintim  7118  nn0supp  9447  uzin  9782  iooval2  10143  fzval2  10239  suprzubdc  10489  bitsinv1  12516  dfphi2  12785  ressabsg  13152  resttopon  14888  restabs  14892  restopnb  14898  txcnmpt  14990