ILE Home Intuitionistic Logic Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  ILE Home  >  Th. List  >  inass GIF version

Theorem inass 3286
Description: Associative law for intersection of classes. Exercise 9 of [TakeutiZaring] p. 17. (Contributed by NM, 3-May-1994.)
Assertion
Ref Expression
inass ((𝐴𝐵) ∩ 𝐶) = (𝐴 ∩ (𝐵𝐶))

Proof of Theorem inass
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 anass 398 . . . 4 (((𝑥𝐴𝑥𝐵) ∧ 𝑥𝐶) ↔ (𝑥𝐴 ∧ (𝑥𝐵𝑥𝐶)))
2 elin 3259 . . . . 5 (𝑥 ∈ (𝐵𝐶) ↔ (𝑥𝐵𝑥𝐶))
32anbi2i 452 . . . 4 ((𝑥𝐴𝑥 ∈ (𝐵𝐶)) ↔ (𝑥𝐴 ∧ (𝑥𝐵𝑥𝐶)))
41, 3bitr4i 186 . . 3 (((𝑥𝐴𝑥𝐵) ∧ 𝑥𝐶) ↔ (𝑥𝐴𝑥 ∈ (𝐵𝐶)))
5 elin 3259 . . . 4 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴𝑥𝐵))
65anbi1i 453 . . 3 ((𝑥 ∈ (𝐴𝐵) ∧ 𝑥𝐶) ↔ ((𝑥𝐴𝑥𝐵) ∧ 𝑥𝐶))
7 elin 3259 . . 3 (𝑥 ∈ (𝐴 ∩ (𝐵𝐶)) ↔ (𝑥𝐴𝑥 ∈ (𝐵𝐶)))
84, 6, 73bitr4i 211 . 2 ((𝑥 ∈ (𝐴𝐵) ∧ 𝑥𝐶) ↔ 𝑥 ∈ (𝐴 ∩ (𝐵𝐶)))
98ineqri 3269 1 ((𝐴𝐵) ∩ 𝐶) = (𝐴 ∩ (𝐵𝐶))
Colors of variables: wff set class
Syntax hints:  wa 103   = wceq 1331  wcel 1480  cin 3070
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 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121
This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-nfc 2270  df-v 2688  df-in 3077
This theorem is referenced by:  in12  3287  in32  3288  in4  3292  indif2  3320  difun1  3336  dfrab3ss  3354  resres  4831  inres  4836  imainrect  4984  restco  12357  restopnb  12364
  Copyright terms: Public domain W3C validator