Theorem inass 3466

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

Assertion

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

Proof of Theorem inass

Dummy variable x is distinct from all other variables.

Step	Hyp	Ref	Expression
1		anass 630	. . . 4 |- (((x e. A /\ x e. B) /\ x e. C) <-> (x e. A /\ (x e. B /\ x e. C)))
2		elin 3220	. . . . 5 |- (x e. (B i^i C) <-> (x e. B /\ x e. C))
3	2	anbi2i 675	. . . 4 |- ((x e. A /\ x e. (B i^i C)) <-> (x e. A /\ (x e. B /\ x e. C)))
4	1, 3	bitr4i 243	. . 3 |- (((x e. A /\ x e. B) /\ x e. C) <-> (x e. A /\ x e. (B i^i C)))
5		elin 3220	. . . 4 |- (x e. (A i^i B) <-> (x e. A /\ x e. B))
6	5	anbi1i 676	. . 3 |- ((x e. (A i^i B) /\ x e. C) <-> ((x e. A /\ x e. B) /\ x e. C))
7		elin 3220	. . 3 |- (x e. (A i^i (B i^i C)) <-> (x e. A /\ x e. (B i^i C)))
8	4, 6, 7	3bitr4i 268	. 2 |- ((x e. (A i^i B) /\ x e. C) <-> x e. (A i^i (B i^i C)))
9	8	ineqri 3450	1 |- ((A i^i B) i^i C) = (A i^i (B i^i C))

Syntax hints:   /\ wa 358   = wceq 1642   e. wcel 1710   i^i cin 3209

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214

This theorem is referenced by:  in12  3467  in32  3468  in4  3472  indif2  3499  difun1  3515  dfrab3ss  3534  dfif4  3674  ssfin  4471  resres  4981  inres  4986