Theorem inass 1650

Description: Associative law for intersection of classes. Exercise 9 of [TakeutiZaring] p. 17.

Assertion

Ref	Expression
inass	|- ((A i^i B) i^i C) = (A i^i (B i^i C))

Proof of Theorem inass

Step	Hyp	Ref	Expression
1		anass 336	. . . 4 |- (((x e. A /\ x e. B) /\ x e. C) <-> (x e. A /\ (x e. B /\ x e. C)))
2		elin 1635	. . . . 5 |- (x e. (B i^i C) <-> (x e. B /\ x e. C))
3	2	anbi2i 367	. . . 4 |- ((x e. A /\ x e. (B i^i C)) <-> (x e. A /\ (x e. B /\ x e. C)))
4	1, 3	bitr4 154	. . 3 |- (((x e. A /\ x e. B) /\ x e. C) <-> (x e. A /\ x e. (B i^i C)))
5		elin 1635	. . . 4 |- (x e. (A i^i B) <-> (x e. A /\ x e. B))
6	5	anbi1i 368	. . 3 |- ((x e. (A i^i B) /\ x e. C) <-> ((x e. A /\ x e. B) /\ x e. C))
7		elin 1635	. . 3 |- (x e. (A i^i (B i^i C)) <-> (x e. A /\ x e. (B i^i C)))
8	4, 6, 7	3bitr4 158	. 2 |- ((x e. (A i^i B) /\ x e. C) <-> x e. (A i^i (B i^i C)))
9	8	ineqri 1637	1 |- ((A i^i B) i^i C) = (A i^i (B i^i C))

Syntax hints:   /\ wa 196   = wceq 1091   e. wcel 1092   i^i cin 1486

This theorem is referenced by:  in12 1651  in23 1652  in4 1653  difun1 1687  onfr 2237  resabs1 2592  resabs2 2593  resdisj 2656  zfregs 3491  chjass 5407  pjoml2 5495  cmcmlem 5500  cmbr3 5509  fh1 5518  fh2 5519  pj3lem1 5658

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-in 1491

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