Proof of Theorem sbthlem3
| Step | Hyp | Ref
| Expression |
| 1 | | sbthlem.1 |
. . . . . 6
⊢ A
∈ V |
| 2 | | sbthlem.2 |
. . . . . 6
⊢ D =
{x∣(x ⊆ A
∧ (g “ (B ∖ (f
“ x))) ⊆ (A ∖ x))} |
| 3 | 1, 2 | sbthlem2 3350 |
. . . . 5
⊢ (ran g
⊆ A → (A ∖ (g
“ (B ∖ (f “ ∪D)))) ⊆ ∪D) |
| 4 | 1, 2 | sbthlem1 3349 |
. . . . 5
⊢ ∪D ⊆ (A
∖ (g “ (B ∖ (f
“ ∪D)))) |
| 5 | 3, 4 | jctil 240 |
. . . 4
⊢ (ran g
⊆ A → (∪D ⊆ (A ∖ (g
“ (B ∖ (f “ ∪D)))) ∧ (A
∖ (g “ (B ∖ (f
“ ∪D))))
⊆ ∪D)) |
| 6 | | eqss 1516 |
. . . 4
⊢ (∪D = (A ∖
(g “ (B ∖ (f
“ ∪D))))
↔ (∪D
⊆ (A ∖ (g “ (B
∖ (f “ ∪D)))) ∧ (A ∖ (g
“ (B ∖ (f “ ∪D)))) ⊆ ∪D)) |
| 7 | 5, 6 | sylibr 175 |
. . 3
⊢ (ran g
⊆ A → ∪D = (A ∖ (g
“ (B ∖ (f “ ∪D))))) |
| 8 | 7 | difeq2d 1588 |
. 2
⊢ (ran g
⊆ A → (A ∖ ∪D) = (A ∖
(A ∖ (g “ (B
∖ (f “ ∪D)))))) |
| 9 | | imassrn 2611 |
. . . 4
⊢ (g
“ (B ∖ (f “ ∪D))) ⊆ ran g |
| 10 | | sstr2 1510 |
. . . 4
⊢ ((g
“ (B ∖ (f “ ∪D))) ⊆ ran g → (ran g
⊆ A → (g “ (B
∖ (f “ ∪D))) ⊆ A)) |
| 11 | 9, 10 | ax-mp 6 |
. . 3
⊢ (ran g
⊆ A → (g “ (B
∖ (f “ ∪D))) ⊆ A) |
| 12 | | dfss4 1667 |
. . 3
⊢ ((g
“ (B ∖ (f “ ∪D))) ⊆ A
↔ (A ∖ (A ∖ (g
“ (B ∖ (f “ ∪D))))) = (g
“ (B ∖ (f “ ∪D)))) |
| 13 | 11, 12 | sylib 173 |
. 2
⊢ (ran g
⊆ A → (A ∖ (A
∖ (g “ (B ∖ (f
“ ∪D)))))
= (g “ (B ∖ (f
“ ∪D)))) |
| 14 | 8, 13 | eqtr2d 1129 |
1
⊢ (ran g
⊆ A → (g “ (B
∖ (f “ ∪D))) = (A ∖ ∪D)) |