Theorem inass 3414

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

Assertion

Ref	Expression
inass	⊢ ((𝐴 ∩ 𝐵) ∩ 𝐶) = (𝐴 ∩ (𝐵 ∩ 𝐶))

Proof of Theorem inass

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		anass 401	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ 𝑥 ∈ 𝐶) ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶)))
2		elin 3387	. . . . 5 ⊢ (𝑥 ∈ (𝐵 ∩ 𝐶) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶))
3	2	anbi2i 457	. . . 4 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ (𝐵 ∩ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶)))
4	1, 3	bitr4i 187	. . 3 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ 𝑥 ∈ 𝐶) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ (𝐵 ∩ 𝐶)))
5		elin 3387	. . . 4 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵))
6	5	anbi1i 458	. . 3 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝑥 ∈ 𝐶) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ 𝑥 ∈ 𝐶))
7		elin 3387	. . 3 ⊢ (𝑥 ∈ (𝐴 ∩ (𝐵 ∩ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ (𝐵 ∩ 𝐶)))
8	4, 6, 7	3bitr4i 212	. 2 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝑥 ∈ 𝐶) ↔ 𝑥 ∈ (𝐴 ∩ (𝐵 ∩ 𝐶)))
9	8	ineqri 3397	1 ⊢ ((𝐴 ∩ 𝐵) ∩ 𝐶) = (𝐴 ∩ (𝐵 ∩ 𝐶))

Syntax hints:   ∧ wa 104   = wceq 1395   ∈ wcel 2200   ∩ cin 3196

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-v 2801  df-in 3203

This theorem is referenced by:  in12  3415  in32  3416  in4  3420  indif2  3448  difun1  3464  dfrab3ss  3482  resres  5017  inres  5022  imainrect  5174  ressinbasd  13115  ressressg  13116  restco  14856  restopnb  14863