Proof of Theorem indifundif
Step | Hyp | Ref
| Expression |
1 | | difindi 4212 |
. 2
⊢ (𝐴 ∖ (𝐵 ∩ 𝐶)) = ((𝐴 ∖ 𝐵) ∪ (𝐴 ∖ 𝐶)) |
2 | | difundir 4211 |
. . . . 5
⊢ (((𝐴 ∩ 𝐵) ∪ (𝐴 ∖ 𝐵)) ∖ 𝐶) = (((𝐴 ∩ 𝐵) ∖ 𝐶) ∪ ((𝐴 ∖ 𝐵) ∖ 𝐶)) |
3 | | inundif 4409 |
. . . . . 6
⊢ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∖ 𝐵)) = 𝐴 |
4 | 3 | difeq1i 4049 |
. . . . 5
⊢ (((𝐴 ∩ 𝐵) ∪ (𝐴 ∖ 𝐵)) ∖ 𝐶) = (𝐴 ∖ 𝐶) |
5 | | uncom 4083 |
. . . . 5
⊢ (((𝐴 ∩ 𝐵) ∖ 𝐶) ∪ ((𝐴 ∖ 𝐵) ∖ 𝐶)) = (((𝐴 ∖ 𝐵) ∖ 𝐶) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶)) |
6 | 2, 4, 5 | 3eqtr3i 2774 |
. . . 4
⊢ (𝐴 ∖ 𝐶) = (((𝐴 ∖ 𝐵) ∖ 𝐶) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶)) |
7 | 6 | uneq2i 4090 |
. . 3
⊢ ((𝐴 ∖ 𝐵) ∪ (𝐴 ∖ 𝐶)) = ((𝐴 ∖ 𝐵) ∪ (((𝐴 ∖ 𝐵) ∖ 𝐶) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶))) |
8 | | unass 4096 |
. . 3
⊢ (((𝐴 ∖ 𝐵) ∪ ((𝐴 ∖ 𝐵) ∖ 𝐶)) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶)) = ((𝐴 ∖ 𝐵) ∪ (((𝐴 ∖ 𝐵) ∖ 𝐶) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶))) |
9 | | undifabs 4408 |
. . . 4
⊢ ((𝐴 ∖ 𝐵) ∪ ((𝐴 ∖ 𝐵) ∖ 𝐶)) = (𝐴 ∖ 𝐵) |
10 | 9 | uneq1i 4089 |
. . 3
⊢ (((𝐴 ∖ 𝐵) ∪ ((𝐴 ∖ 𝐵) ∖ 𝐶)) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶)) = ((𝐴 ∖ 𝐵) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶)) |
11 | 7, 8, 10 | 3eqtr2i 2772 |
. 2
⊢ ((𝐴 ∖ 𝐵) ∪ (𝐴 ∖ 𝐶)) = ((𝐴 ∖ 𝐵) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶)) |
12 | | uncom 4083 |
. 2
⊢ ((𝐴 ∖ 𝐵) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶)) = (((𝐴 ∩ 𝐵) ∖ 𝐶) ∪ (𝐴 ∖ 𝐵)) |
13 | 1, 11, 12 | 3eqtrri 2771 |
1
⊢ (((𝐴 ∩ 𝐵) ∖ 𝐶) ∪ (𝐴 ∖ 𝐵)) = (𝐴 ∖ (𝐵 ∩ 𝐶)) |