Theorem inass 3384

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 3357	. . . . 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 3357	. . . 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 3357	. . 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 3367	1 ⊢ ((𝐴 ∩ 𝐵) ∩ 𝐶) = (𝐴 ∩ (𝐵 ∩ 𝐶))

Syntax hints:   ∧ wa 104   = wceq 1373   ∈ wcel 2177   ∩ cin 3166

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 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2188

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-nfc 2338  df-v 2775  df-in 3173

This theorem is referenced by:  in12  3385  in32  3386  in4  3390  indif2  3418  difun1  3434  dfrab3ss  3452  resres  4976  inres  4981  imainrect  5133  ressinbasd  12950  ressressg  12951  restco  14690  restopnb  14697