Theorem inass 3281

Description: Associative law for intersection of classes. Exercise 9 of [TakeutiZaring] p. 17. (Contributed by NM, 3-May-1994.)

Assertion

Ref	Expression
inass	⊢ ((𝐴 ∩ 𝐵) ∩ 𝐶) = (𝐴 ∩ (𝐵 ∩ 𝐶))

Proof of Theorem inass

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		anass 398	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ 𝑥 ∈ 𝐶) ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶)))
2		elin 3254	. . . . 5 ⊢ (𝑥 ∈ (𝐵 ∩ 𝐶) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶))
3	2	anbi2i 452	. . . 4 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ (𝐵 ∩ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶)))
4	1, 3	bitr4i 186	. . 3 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ 𝑥 ∈ 𝐶) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ (𝐵 ∩ 𝐶)))
5		elin 3254	. . . 4 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵))
6	5	anbi1i 453	. . 3 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝑥 ∈ 𝐶) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ 𝑥 ∈ 𝐶))
7		elin 3254	. . 3 ⊢ (𝑥 ∈ (𝐴 ∩ (𝐵 ∩ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ (𝐵 ∩ 𝐶)))
8	4, 6, 7	3bitr4i 211	. 2 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝑥 ∈ 𝐶) ↔ 𝑥 ∈ (𝐴 ∩ (𝐵 ∩ 𝐶)))
9	8	ineqri 3264	1 ⊢ ((𝐴 ∩ 𝐵) ∩ 𝐶) = (𝐴 ∩ (𝐵 ∩ 𝐶))

Syntax hints:   ∧ wa 103   = wceq 1331   ∈ wcel 1480   ∩ cin 3065

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 2119

This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2124  df-cleq 2130  df-clel 2133  df-nfc 2268  df-v 2683  df-in 3072

This theorem is referenced by:  in12  3282  in32  3283  in4  3287  indif2  3315  difun1  3331  dfrab3ss  3349  resres  4826  inres  4831  imainrect  4979  restco  12332  restopnb  12339