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 ( ( 𝐴𝐵 ) ∩ 𝐶 ) = ( 𝐴 ∩ ( 𝐵𝐶 ) )

Proof

Step Hyp Ref Expression
1 anass ( ( ( 𝑥𝐴𝑥𝐵 ) ∧ 𝑥𝐶 ) ↔ ( 𝑥𝐴 ∧ ( 𝑥𝐵𝑥𝐶 ) ) )
2 elin ( 𝑥 ∈ ( 𝐵𝐶 ) ↔ ( 𝑥𝐵𝑥𝐶 ) )
3 2 anbi2i ( ( 𝑥𝐴𝑥 ∈ ( 𝐵𝐶 ) ) ↔ ( 𝑥𝐴 ∧ ( 𝑥𝐵𝑥𝐶 ) ) )
4 1 3 bitr4i ( ( ( 𝑥𝐴𝑥𝐵 ) ∧ 𝑥𝐶 ) ↔ ( 𝑥𝐴𝑥 ∈ ( 𝐵𝐶 ) ) )
5 elin ( 𝑥 ∈ ( 𝐴𝐵 ) ↔ ( 𝑥𝐴𝑥𝐵 ) )
6 5 anbi1i ( ( 𝑥 ∈ ( 𝐴𝐵 ) ∧ 𝑥𝐶 ) ↔ ( ( 𝑥𝐴𝑥𝐵 ) ∧ 𝑥𝐶 ) )
7 elin ( 𝑥 ∈ ( 𝐴 ∩ ( 𝐵𝐶 ) ) ↔ ( 𝑥𝐴𝑥 ∈ ( 𝐵𝐶 ) ) )
8 4 6 7 3bitr4i ( ( 𝑥 ∈ ( 𝐴𝐵 ) ∧ 𝑥𝐶 ) ↔ 𝑥 ∈ ( 𝐴 ∩ ( 𝐵𝐶 ) ) )
9 8 ineqri ( ( 𝐴𝐵 ) ∩ 𝐶 ) = ( 𝐴 ∩ ( 𝐵𝐶 ) )