Theorem rp-fakeuninass 43961

Description: A special case where a mixture of union and intersection appears to conform to a mixed associative law. (Contributed by RP, 29-Feb-2020.)

Assertion

Ref	Expression
rp-fakeuninass	⊢ (𝐴 ⊆ 𝐶 ↔ ((𝐴 ∪ 𝐵) ∩ 𝐶) = (𝐴 ∪ (𝐵 ∩ 𝐶)))

Proof of Theorem rp-fakeuninass

Step	Hyp	Ref	Expression
1		rp-fakeinunass 43960	. 2 ⊢ (𝐴 ⊆ 𝐶 ↔ ((𝐶 ∩ 𝐵) ∪ 𝐴) = (𝐶 ∩ (𝐵 ∪ 𝐴)))
2		eqcom 2747	. 2 ⊢ (((𝐶 ∩ 𝐵) ∪ 𝐴) = (𝐶 ∩ (𝐵 ∪ 𝐴)) ↔ (𝐶 ∩ (𝐵 ∪ 𝐴)) = ((𝐶 ∩ 𝐵) ∪ 𝐴))
3		incom 4145	. . . 4 ⊢ (𝐶 ∩ (𝐵 ∪ 𝐴)) = ((𝐵 ∪ 𝐴) ∩ 𝐶)
4		uncom 4095	. . . . 5 ⊢ (𝐵 ∪ 𝐴) = (𝐴 ∪ 𝐵)
5	4	ineq1i 4152	. . . 4 ⊢ ((𝐵 ∪ 𝐴) ∩ 𝐶) = ((𝐴 ∪ 𝐵) ∩ 𝐶)
6	3, 5	eqtri 2763	. . 3 ⊢ (𝐶 ∩ (𝐵 ∪ 𝐴)) = ((𝐴 ∪ 𝐵) ∩ 𝐶)
7		uncom 4095	. . . 4 ⊢ ((𝐶 ∩ 𝐵) ∪ 𝐴) = (𝐴 ∪ (𝐶 ∩ 𝐵))
8		incom 4145	. . . . 5 ⊢ (𝐶 ∩ 𝐵) = (𝐵 ∩ 𝐶)
9	8	uneq2i 4102	. . . 4 ⊢ (𝐴 ∪ (𝐶 ∩ 𝐵)) = (𝐴 ∪ (𝐵 ∩ 𝐶))
10	7, 9	eqtri 2763	. . 3 ⊢ ((𝐶 ∩ 𝐵) ∪ 𝐴) = (𝐴 ∪ (𝐵 ∩ 𝐶))
11	6, 10	eqeq12i 2758	. 2 ⊢ ((𝐶 ∩ (𝐵 ∪ 𝐴)) = ((𝐶 ∩ 𝐵) ∪ 𝐴) ↔ ((𝐴 ∪ 𝐵) ∩ 𝐶) = (𝐴 ∪ (𝐵 ∩ 𝐶)))
12	1, 2, 11	3bitri 298	1 ⊢ (𝐴 ⊆ 𝐶 ↔ ((𝐴 ∪ 𝐵) ∩ 𝐶) = (𝐴 ∪ (𝐵 ∩ 𝐶)))

Syntax hints:   ↔ wb 207   = wceq 1547   ∪ cun 3888   ∩ cin 3889   ⊆ wss 3890

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2712

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-tru 1550  df-ex 1787  df-sb 2074  df-clab 2719  df-cleq 2732  df-clel 2815  df-rab 3393  df-v 3434  df-un 3895  df-in 3897  df-ss 3907

This theorem is referenced by: (None)