# Metamath Proof Explorer

## Theorem inass

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

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

### Proof

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