Theorem inass 3369

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 3342	. . . . 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 3342	. . . 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 3342	. . 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 3352	1 ⊢ ((𝐴 ∩ 𝐵) ∩ 𝐶) = (𝐴 ∩ (𝐵 ∩ 𝐶))

Syntax hints:   ∧ wa 104   = wceq 1364   ∈ wcel 2164   ∩ cin 3152

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 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2175

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-v 2762  df-in 3159

This theorem is referenced by:  in12  3370  in32  3371  in4  3375  indif2  3403  difun1  3419  dfrab3ss  3437  resres  4954  inres  4959  imainrect  5111  ressinbasd  12692  ressressg  12693  restco  14342  restopnb  14349