Proof of Theorem difin
| Step | Hyp | Ref
| Expression |
| 1 | | abai 366 |
. . . 4
⊢ ((x
∈ A ∧ ¬ x ∈ B)
↔ (x ∈ A ∧ (x
∈ A → ¬ x ∈ B))) |
| 2 | | imnan 207 |
. . . . 5
⊢ ((x
∈ A → ¬ x ∈ B)
↔ ¬ (x ∈ A ∧ x ∈
B)) |
| 3 | 2 | anbi2i 367 |
. . . 4
⊢ ((x
∈ A ∧ (x ∈ A
→ ¬ x ∈ B)) ↔ (x
∈ A ∧ ¬ (x ∈ A ∧
x ∈ B))) |
| 4 | 1, 3 | bitr 151 |
. . 3
⊢ ((x
∈ A ∧ ¬ x ∈ B)
↔ (x ∈ A ∧ ¬ (x
∈ A ∧ x ∈ B))) |
| 5 | | eldif 1496 |
. . 3
⊢ (x
∈ (A ∖ B) ↔ (x
∈ A ∧ ¬ x ∈ B)) |
| 6 | | eldif 1496 |
. . . 4
⊢ (x
∈ (A ∖ (A ∩ B))
↔ (x ∈ A ∧ ¬ x
∈ (A ∩ B))) |
| 7 | | elin 1635 |
. . . . . 6
⊢ (x
∈ (A ∩ B) ↔ (x
∈ A ∧ x ∈ B)) |
| 8 | 7 | negbii 162 |
. . . . 5
⊢ (¬ x ∈ (A
∩ B) ↔ ¬ (x ∈ A ∧
x ∈ B)) |
| 9 | 8 | anbi2i 367 |
. . . 4
⊢ ((x
∈ A ∧ ¬ x ∈ (A
∩ B)) ↔ (x ∈ A ∧
¬ (x ∈ A ∧ x ∈
B))) |
| 10 | 6, 9 | bitr 151 |
. . 3
⊢ (x
∈ (A ∖ (A ∩ B))
↔ (x ∈ A ∧ ¬ (x
∈ A ∧ x ∈ B))) |
| 11 | 4, 5, 10 | 3bitr4r 159 |
. 2
⊢ (x
∈ (A ∖ (A ∩ B))
↔ x ∈ (A ∖ B)) |
| 12 | 11 | cleqri 1101 |
1
⊢ (A
∖ (A ∩ B)) = (A ∖
B) |