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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 ) )