Theorem rp-fakeuninass 44709

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 44708	. 2 ⊢ (𝐴 ⊆ 𝐶 ↔ ((𝐶 ∩ 𝐵) ∪ 𝐴) = (𝐶 ∩ (𝐵 ∪ 𝐴)))
2		eqcom 2770	. 2 ⊢ (((𝐶 ∩ 𝐵) ∪ 𝐴) = (𝐶 ∩ (𝐵 ∪ 𝐴)) ↔ (𝐶 ∩ (𝐵 ∪ 𝐴)) = ((𝐶 ∩ 𝐵) ∪ 𝐴))
3		incom 4109	. . . 4 ⊢ (𝐶 ∩ (𝐵 ∪ 𝐴)) = ((𝐵 ∪ 𝐴) ∩ 𝐶)
4		uncom 4160	. . . . 5 ⊢ (𝐵 ∪ 𝐴) = (𝐴 ∪ 𝐵)
5	4	ineq1i 4116	. . . 4 ⊢ ((𝐵 ∪ 𝐴) ∩ 𝐶) = ((𝐴 ∪ 𝐵) ∩ 𝐶)
6	3, 5	eqtri 2786	. . 3 ⊢ (𝐶 ∩ (𝐵 ∪ 𝐴)) = ((𝐴 ∪ 𝐵) ∩ 𝐶)
7		uncom 4160	. . . 4 ⊢ ((𝐶 ∩ 𝐵) ∪ 𝐴) = (𝐴 ∪ (𝐶 ∩ 𝐵))
8		incom 4109	. . . . 5 ⊢ (𝐶 ∩ 𝐵) = (𝐵 ∩ 𝐶)
9	8	uneq2i 4167	. . . 4 ⊢ (𝐴 ∪ (𝐶 ∩ 𝐵)) = (𝐴 ∪ (𝐵 ∩ 𝐶))
10	7, 9	eqtri 2786	. . 3 ⊢ ((𝐶 ∩ 𝐵) ∪ 𝐴) = (𝐴 ∪ (𝐵 ∩ 𝐶))
11	6, 10	eqeq12i 2781	. 2 ⊢ ((𝐶 ∩ (𝐵 ∪ 𝐴)) = ((𝐶 ∩ 𝐵) ∪ 𝐴) ↔ ((𝐴 ∪ 𝐵) ∩ 𝐶) = (𝐴 ∪ (𝐵 ∩ 𝐶)))
12	1, 2, 11	3bitri 300	1 ⊢ (𝐴 ⊆ 𝐶 ↔ ((𝐴 ∪ 𝐵) ∩ 𝐶) = (𝐴 ∪ (𝐵 ∩ 𝐶)))

This proof depends on syntax axioms:   ↔ wb 209   = wceq 1570   ⊆ wss 3900   ∩ cin 4092   ∪ cun 4152

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-rab 3417  df-v 3456  df-ss 3901  df-in 4094  df-un 4154

This theorem is used by: (None)