Theorem iundifdifd 29718

Description: The intersection of a set is the complement of the union of the complements. (Contributed by Thierry Arnoux, 19-Dec-2016.)

Assertion

Ref	Expression
iundifdifd	⊢ (𝐴 ⊆ 𝒫 𝑂 → (𝐴 ≠ ∅ → ∩ 𝐴 = (𝑂 ∖ ∪ 𝑥 ∈ 𝐴 (𝑂 ∖ 𝑥))))

Distinct variable groups:   𝑥,𝐴   𝑥,𝑂

Proof of Theorem iundifdifd

Step	Hyp	Ref	Expression
1		iundif2 4721	. . . . 5 ⊢ ∪ 𝑥 ∈ 𝐴 (𝑂 ∖ 𝑥) = (𝑂 ∖ ∩ 𝑥 ∈ 𝐴 𝑥)
2		intiin 4708	. . . . . 6 ⊢ ∩ 𝐴 = ∩ 𝑥 ∈ 𝐴 𝑥
3	2	difeq2i 3876	. . . . 5 ⊢ (𝑂 ∖ ∩ 𝐴) = (𝑂 ∖ ∩ 𝑥 ∈ 𝐴 𝑥)
4	1, 3	eqtr4i 2796	. . . 4 ⊢ ∪ 𝑥 ∈ 𝐴 (𝑂 ∖ 𝑥) = (𝑂 ∖ ∩ 𝐴)
5	4	difeq2i 3876	. . 3 ⊢ (𝑂 ∖ ∪ 𝑥 ∈ 𝐴 (𝑂 ∖ 𝑥)) = (𝑂 ∖ (𝑂 ∖ ∩ 𝐴))
6		intssuni2 4636	. . . . 5 ⊢ ((𝐴 ⊆ 𝒫 𝑂 ∧ 𝐴 ≠ ∅) → ∩ 𝐴 ⊆ ∪ 𝒫 𝑂)
7		unipw 5046	. . . . 5 ⊢ ∪ 𝒫 𝑂 = 𝑂
8	6, 7	syl6sseq 3800	. . . 4 ⊢ ((𝐴 ⊆ 𝒫 𝑂 ∧ 𝐴 ≠ ∅) → ∩ 𝐴 ⊆ 𝑂)
9		dfss4 4007	. . . 4 ⊢ (∩ 𝐴 ⊆ 𝑂 ↔ (𝑂 ∖ (𝑂 ∖ ∩ 𝐴)) = ∩ 𝐴)
10	8, 9	sylib 208	. . 3 ⊢ ((𝐴 ⊆ 𝒫 𝑂 ∧ 𝐴 ≠ ∅) → (𝑂 ∖ (𝑂 ∖ ∩ 𝐴)) = ∩ 𝐴)
11	5, 10	syl5req 2818	. 2 ⊢ ((𝐴 ⊆ 𝒫 𝑂 ∧ 𝐴 ≠ ∅) → ∩ 𝐴 = (𝑂 ∖ ∪ 𝑥 ∈ 𝐴 (𝑂 ∖ 𝑥)))
12	11	ex 397	1 ⊢ (𝐴 ⊆ 𝒫 𝑂 → (𝐴 ≠ ∅ → ∩ 𝐴 = (𝑂 ∖ ∪ 𝑥 ∈ 𝐴 (𝑂 ∖ 𝑥))))

Syntax hints:   → wi 4   ∧ wa 382   = wceq 1631   ≠ wne 2943   ∖ cdif 3720   ⊆ wss 3723  ∅c0 4063  𝒫 cpw 4297  ∪ cuni 4574  ∩ cint 4611  ∪ ciun 4654  ∩ ciin 4655

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1870  ax-4 1885  ax-5 1991  ax-6 2057  ax-7 2093  ax-9 2154  ax-10 2174  ax-11 2190  ax-12 2203  ax-13 2408  ax-ext 2751  ax-sep 4915  ax-nul 4923  ax-pr 5034

This theorem depends on definitions:  df-bi 197  df-an 383  df-or 827  df-3an 1073  df-tru 1634  df-ex 1853  df-nf 1858  df-sb 2050  df-clab 2758  df-cleq 2764  df-clel 2767  df-nfc 2902  df-ne 2944  df-ral 3066  df-rex 3067  df-rab 3070  df-v 3353  df-dif 3726  df-un 3728  df-in 3730  df-ss 3737  df-nul 4064  df-pw 4299  df-sn 4317  df-pr 4319  df-uni 4575  df-int 4612  df-iun 4656  df-iin 4657

This theorem is referenced by:  sigaclci  30535