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Theorem inass 3120
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 383 . . . 4 (((x A x B) x 𝐶) ↔ (x A (x B x 𝐶)))
2 elin 3099 . . . . 5 (x (B𝐶) ↔ (x B x 𝐶))
32anbi2i 433 . . . 4 ((x A x (B𝐶)) ↔ (x A (x B x 𝐶)))
41, 3bitr4i 176 . . 3 (((x A x B) x 𝐶) ↔ (x A x (B𝐶)))
5 elin 3099 . . . 4 (x (AB) ↔ (x A x B))
65anbi1i 434 . . 3 ((x (AB) x 𝐶) ↔ ((x A x B) x 𝐶))
7 elin 3099 . . 3 (x (A ∩ (B𝐶)) ↔ (x A x (B𝐶)))
84, 6, 73bitr4i 201 . 2 ((x (AB) x 𝐶) ↔ x (A ∩ (B𝐶)))
98ineqri 3103 1 ((AB) ∩ 𝐶) = (A ∩ (B𝐶))
Colors of variables: wff set class
Syntax hints:   wa 97   = wceq 1226   wcel 1370  cin 2889
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
This theorem is referenced by:  in12  3121  in32  3122  in4  3126  indif2  3154  difun1  3170  dfrab3ss  3188  resres  4547  inres  4552  imainrect  4689
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