Proof of Theorem difeq
Step | Hyp | Ref
| Expression |
1 | | ineq1 4136 |
. . . 4
⊢ ((𝐴 ∖ 𝐵) = 𝐶 → ((𝐴 ∖ 𝐵) ∩ 𝐵) = (𝐶 ∩ 𝐵)) |
2 | | disjdifr 4403 |
. . . 4
⊢ ((𝐴 ∖ 𝐵) ∩ 𝐵) = ∅ |
3 | 1, 2 | eqtr3di 2794 |
. . 3
⊢ ((𝐴 ∖ 𝐵) = 𝐶 → (𝐶 ∩ 𝐵) = ∅) |
4 | | uneq1 4086 |
. . . 4
⊢ ((𝐴 ∖ 𝐵) = 𝐶 → ((𝐴 ∖ 𝐵) ∪ 𝐵) = (𝐶 ∪ 𝐵)) |
5 | | undif1 4406 |
. . . 4
⊢ ((𝐴 ∖ 𝐵) ∪ 𝐵) = (𝐴 ∪ 𝐵) |
6 | 4, 5 | eqtr3di 2794 |
. . 3
⊢ ((𝐴 ∖ 𝐵) = 𝐶 → (𝐶 ∪ 𝐵) = (𝐴 ∪ 𝐵)) |
7 | 3, 6 | jca 511 |
. 2
⊢ ((𝐴 ∖ 𝐵) = 𝐶 → ((𝐶 ∩ 𝐵) = ∅ ∧ (𝐶 ∪ 𝐵) = (𝐴 ∪ 𝐵))) |
8 | | simpl 482 |
. . . 4
⊢ (((𝐶 ∩ 𝐵) = ∅ ∧ (𝐶 ∪ 𝐵) = (𝐴 ∪ 𝐵)) → (𝐶 ∩ 𝐵) = ∅) |
9 | | disj3 4384 |
. . . . 5
⊢ ((𝐶 ∩ 𝐵) = ∅ ↔ 𝐶 = (𝐶 ∖ 𝐵)) |
10 | | eqcom 2745 |
. . . . 5
⊢ (𝐶 = (𝐶 ∖ 𝐵) ↔ (𝐶 ∖ 𝐵) = 𝐶) |
11 | 9, 10 | bitri 274 |
. . . 4
⊢ ((𝐶 ∩ 𝐵) = ∅ ↔ (𝐶 ∖ 𝐵) = 𝐶) |
12 | 8, 11 | sylib 217 |
. . 3
⊢ (((𝐶 ∩ 𝐵) = ∅ ∧ (𝐶 ∪ 𝐵) = (𝐴 ∪ 𝐵)) → (𝐶 ∖ 𝐵) = 𝐶) |
13 | | difeq1 4046 |
. . . . . 6
⊢ ((𝐶 ∪ 𝐵) = (𝐴 ∪ 𝐵) → ((𝐶 ∪ 𝐵) ∖ 𝐵) = ((𝐴 ∪ 𝐵) ∖ 𝐵)) |
14 | | difun2 4411 |
. . . . . 6
⊢ ((𝐶 ∪ 𝐵) ∖ 𝐵) = (𝐶 ∖ 𝐵) |
15 | | difun2 4411 |
. . . . . 6
⊢ ((𝐴 ∪ 𝐵) ∖ 𝐵) = (𝐴 ∖ 𝐵) |
16 | 13, 14, 15 | 3eqtr3g 2802 |
. . . . 5
⊢ ((𝐶 ∪ 𝐵) = (𝐴 ∪ 𝐵) → (𝐶 ∖ 𝐵) = (𝐴 ∖ 𝐵)) |
17 | 16 | eqeq1d 2740 |
. . . 4
⊢ ((𝐶 ∪ 𝐵) = (𝐴 ∪ 𝐵) → ((𝐶 ∖ 𝐵) = 𝐶 ↔ (𝐴 ∖ 𝐵) = 𝐶)) |
18 | 17 | adantl 481 |
. . 3
⊢ (((𝐶 ∩ 𝐵) = ∅ ∧ (𝐶 ∪ 𝐵) = (𝐴 ∪ 𝐵)) → ((𝐶 ∖ 𝐵) = 𝐶 ↔ (𝐴 ∖ 𝐵) = 𝐶)) |
19 | 12, 18 | mpbid 231 |
. 2
⊢ (((𝐶 ∩ 𝐵) = ∅ ∧ (𝐶 ∪ 𝐵) = (𝐴 ∪ 𝐵)) → (𝐴 ∖ 𝐵) = 𝐶) |
20 | 7, 19 | impbii 208 |
1
⊢ ((𝐴 ∖ 𝐵) = 𝐶 ↔ ((𝐶 ∩ 𝐵) = ∅ ∧ (𝐶 ∪ 𝐵) = (𝐴 ∪ 𝐵))) |