Proof of Theorem uneqdifeqim
Step | Hyp | Ref
| Expression |
1 | | uncom 3266 |
. . . 4
⊢ (𝐵 ∪ 𝐴) = (𝐴 ∪ 𝐵) |
2 | | eqtr 2183 |
. . . . . 6
⊢ (((𝐵 ∪ 𝐴) = (𝐴 ∪ 𝐵) ∧ (𝐴 ∪ 𝐵) = 𝐶) → (𝐵 ∪ 𝐴) = 𝐶) |
3 | 2 | eqcomd 2171 |
. . . . 5
⊢ (((𝐵 ∪ 𝐴) = (𝐴 ∪ 𝐵) ∧ (𝐴 ∪ 𝐵) = 𝐶) → 𝐶 = (𝐵 ∪ 𝐴)) |
4 | | difeq1 3233 |
. . . . . 6
⊢ (𝐶 = (𝐵 ∪ 𝐴) → (𝐶 ∖ 𝐴) = ((𝐵 ∪ 𝐴) ∖ 𝐴)) |
5 | | difun2 3488 |
. . . . . 6
⊢ ((𝐵 ∪ 𝐴) ∖ 𝐴) = (𝐵 ∖ 𝐴) |
6 | | eqtr 2183 |
. . . . . . 7
⊢ (((𝐶 ∖ 𝐴) = ((𝐵 ∪ 𝐴) ∖ 𝐴) ∧ ((𝐵 ∪ 𝐴) ∖ 𝐴) = (𝐵 ∖ 𝐴)) → (𝐶 ∖ 𝐴) = (𝐵 ∖ 𝐴)) |
7 | | incom 3314 |
. . . . . . . . . 10
⊢ (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐴) |
8 | 7 | eqeq1i 2173 |
. . . . . . . . 9
⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ (𝐵 ∩ 𝐴) = ∅) |
9 | | disj3 3461 |
. . . . . . . . 9
⊢ ((𝐵 ∩ 𝐴) = ∅ ↔ 𝐵 = (𝐵 ∖ 𝐴)) |
10 | 8, 9 | bitri 183 |
. . . . . . . 8
⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐵 = (𝐵 ∖ 𝐴)) |
11 | | eqtr 2183 |
. . . . . . . . . 10
⊢ (((𝐶 ∖ 𝐴) = (𝐵 ∖ 𝐴) ∧ (𝐵 ∖ 𝐴) = 𝐵) → (𝐶 ∖ 𝐴) = 𝐵) |
12 | 11 | expcom 115 |
. . . . . . . . 9
⊢ ((𝐵 ∖ 𝐴) = 𝐵 → ((𝐶 ∖ 𝐴) = (𝐵 ∖ 𝐴) → (𝐶 ∖ 𝐴) = 𝐵)) |
13 | 12 | eqcoms 2168 |
. . . . . . . 8
⊢ (𝐵 = (𝐵 ∖ 𝐴) → ((𝐶 ∖ 𝐴) = (𝐵 ∖ 𝐴) → (𝐶 ∖ 𝐴) = 𝐵)) |
14 | 10, 13 | sylbi 120 |
. . . . . . 7
⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐶 ∖ 𝐴) = (𝐵 ∖ 𝐴) → (𝐶 ∖ 𝐴) = 𝐵)) |
15 | 6, 14 | syl5com 29 |
. . . . . 6
⊢ (((𝐶 ∖ 𝐴) = ((𝐵 ∪ 𝐴) ∖ 𝐴) ∧ ((𝐵 ∪ 𝐴) ∖ 𝐴) = (𝐵 ∖ 𝐴)) → ((𝐴 ∩ 𝐵) = ∅ → (𝐶 ∖ 𝐴) = 𝐵)) |
16 | 4, 5, 15 | sylancl 410 |
. . . . 5
⊢ (𝐶 = (𝐵 ∪ 𝐴) → ((𝐴 ∩ 𝐵) = ∅ → (𝐶 ∖ 𝐴) = 𝐵)) |
17 | 3, 16 | syl 14 |
. . . 4
⊢ (((𝐵 ∪ 𝐴) = (𝐴 ∪ 𝐵) ∧ (𝐴 ∪ 𝐵) = 𝐶) → ((𝐴 ∩ 𝐵) = ∅ → (𝐶 ∖ 𝐴) = 𝐵)) |
18 | 1, 17 | mpan 421 |
. . 3
⊢ ((𝐴 ∪ 𝐵) = 𝐶 → ((𝐴 ∩ 𝐵) = ∅ → (𝐶 ∖ 𝐴) = 𝐵)) |
19 | 18 | com12 30 |
. 2
⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∪ 𝐵) = 𝐶 → (𝐶 ∖ 𝐴) = 𝐵)) |
20 | 19 | adantl 275 |
1
⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐴 ∩ 𝐵) = ∅) → ((𝐴 ∪ 𝐵) = 𝐶 → (𝐶 ∖ 𝐴) = 𝐵)) |