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Theorem inss 3160
 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 3159 . 2 (A𝐶 → (AB) ⊆ 𝐶)
2 incom 3123 . . 3 (AB) = (BA)
3 ssinss1 3159 . . 3 (B𝐶 → (BA) ⊆ 𝐶)
42, 3syl5eqss 2983 . 2 (B𝐶 → (AB) ⊆ 𝐶)
51, 4jaoi 635 1 ((A𝐶 B𝐶) → (AB) ⊆ 𝐶)
 Colors of variables: wff set class Syntax hints:   → wi 4   ∨ wo 628   ∩ cin 2910   ⊆ wss 2911 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 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019 This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-v 2553  df-in 2918  df-ss 2925 This theorem is referenced by: (None)
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