Theorem inass 3291

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 399	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ 𝑥 ∈ 𝐶) ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶)))
2		elin 3264	. . . . 5 ⊢ (𝑥 ∈ (𝐵 ∩ 𝐶) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶))
3	2	anbi2i 453	. . . 4 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ (𝐵 ∩ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶)))
4	1, 3	bitr4i 186	. . 3 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ 𝑥 ∈ 𝐶) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ (𝐵 ∩ 𝐶)))
5		elin 3264	. . . 4 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵))
6	5	anbi1i 454	. . 3 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝑥 ∈ 𝐶) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ 𝑥 ∈ 𝐶))
7		elin 3264	. . 3 ⊢ (𝑥 ∈ (𝐴 ∩ (𝐵 ∩ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ (𝐵 ∩ 𝐶)))
8	4, 6, 7	3bitr4i 211	. 2 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝑥 ∈ 𝐶) ↔ 𝑥 ∈ (𝐴 ∩ (𝐵 ∩ 𝐶)))
9	8	ineqri 3274	1 ⊢ ((𝐴 ∩ 𝐵) ∩ 𝐶) = (𝐴 ∩ (𝐵 ∩ 𝐶))

Syntax hints:   ∧ wa 103   = wceq 1332   ∈ wcel 1481   ∩ cin 3075

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-bndl 1487  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516  ax-ext 2122

This theorem depends on definitions:  df-bi 116  df-tru 1335  df-nf 1438  df-sb 1737  df-clab 2127  df-cleq 2133  df-clel 2136  df-nfc 2271  df-v 2691  df-in 3082

This theorem is referenced by:  in12  3292  in32  3293  in4  3297  indif2  3325  difun1  3341  dfrab3ss  3359  resres  4839  inres  4844  imainrect  4992  restco  12382  restopnb  12389