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Theorem inss 3139
Description: Inclusion of an intersection of two classes. (Contributed by NM, 30-Oct-2014.)
Assertion
Ref Expression
inss ((A𝐶 B𝐶) → (AB) ⊆ 𝐶)

Proof of Theorem inss
StepHypRef Expression
1 ssinss1 3138 . 2 (A𝐶 → (AB) ⊆ 𝐶)
2 incom 3102 . . 3 (AB) = (BA)
3 ssinss1 3138 . . 3 (B𝐶 → (BA) ⊆ 𝐶)
42, 3syl5eqss 2962 . 2 (B𝐶 → (AB) ⊆ 𝐶)
51, 4jaoi 623 1 ((A𝐶 B𝐶) → (AB) ⊆ 𝐶)
Colors of variables: wff set class
Syntax hints:  wi 4   wo 616  cin 2889  wss 2890
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 617  ax-5 1312  ax-7 1313  ax-gen 1314  ax-ie1 1359  ax-ie2 1360  ax-8 1372  ax-10 1373  ax-11 1374  ax-i12 1375  ax-bnd 1376  ax-4 1377  ax-17 1396  ax-i9 1400  ax-ial 1405  ax-i5r 1406  ax-ext 2000
This theorem depends on definitions:  df-bi 110  df-tru 1229  df-nf 1326  df-sb 1624  df-clab 2005  df-cleq 2011  df-clel 2014  df-nfc 2145  df-v 2533  df-in 2897  df-ss 2904
This theorem is referenced by: (None)
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