Proof of Theorem sbthlem6
| Step | Hyp | Ref
| Expression |
| 1 | | rnun 6100 |
. . . 4
⊢ ran
((𝑓 ↾ ∪ 𝐷)
∪ (◡𝑔 ↾ (𝐴 ∖ ∪ 𝐷))) = (ran (𝑓 ↾ ∪ 𝐷) ∪ ran (◡𝑔 ↾ (𝐴 ∖ ∪ 𝐷))) |
| 2 | | sbthlem.3 |
. . . . 5
⊢ 𝐻 = ((𝑓 ↾ ∪ 𝐷) ∪ (◡𝑔 ↾ (𝐴 ∖ ∪ 𝐷))) |
| 3 | 2 | rneqi 5883 |
. . . 4
⊢ ran 𝐻 = ran ((𝑓 ↾ ∪ 𝐷) ∪ (◡𝑔 ↾ (𝐴 ∖ ∪ 𝐷))) |
| 4 | | df-ima 5634 |
. . . . 5
⊢ (𝑓 “ ∪ 𝐷) =
ran (𝑓 ↾ ∪ 𝐷) |
| 5 | 4 | uneq1i 4113 |
. . . 4
⊢ ((𝑓 “ ∪ 𝐷)
∪ ran (◡𝑔 ↾ (𝐴 ∖ ∪ 𝐷))) = (ran (𝑓 ↾ ∪ 𝐷) ∪ ran (◡𝑔 ↾ (𝐴 ∖ ∪ 𝐷))) |
| 6 | 1, 3, 5 | 3eqtr4i 2766 |
. . 3
⊢ ran 𝐻 = ((𝑓 “ ∪ 𝐷) ∪ ran (◡𝑔 ↾ (𝐴 ∖ ∪ 𝐷))) |
| 7 | | sbthlem.1 |
. . . . . 6
⊢ 𝐴 ∈ V |
| 8 | | sbthlem.2 |
. . . . . 6
⊢ 𝐷 = {𝑥 ∣ (𝑥 ⊆ 𝐴 ∧ (𝑔 “ (𝐵 ∖ (𝑓 “ 𝑥))) ⊆ (𝐴 ∖ 𝑥))} |
| 9 | 7, 8 | sbthlem4 9014 |
. . . . 5
⊢ (((dom
𝑔 = 𝐵 ∧ ran 𝑔 ⊆ 𝐴) ∧ Fun ◡𝑔) → (◡𝑔 “ (𝐴 ∖ ∪ 𝐷)) = (𝐵 ∖ (𝑓 “ ∪ 𝐷))) |
| 10 | | df-ima 5634 |
. . . . 5
⊢ (◡𝑔 “ (𝐴 ∖ ∪ 𝐷)) = ran (◡𝑔 ↾ (𝐴 ∖ ∪ 𝐷)) |
| 11 | 9, 10 | eqtr3di 2783 |
. . . 4
⊢ (((dom
𝑔 = 𝐵 ∧ ran 𝑔 ⊆ 𝐴) ∧ Fun ◡𝑔) → (𝐵 ∖ (𝑓 “ ∪ 𝐷)) = ran (◡𝑔 ↾ (𝐴 ∖ ∪ 𝐷))) |
| 12 | 11 | uneq2d 4117 |
. . 3
⊢ (((dom
𝑔 = 𝐵 ∧ ran 𝑔 ⊆ 𝐴) ∧ Fun ◡𝑔) → ((𝑓 “ ∪ 𝐷) ∪ (𝐵 ∖ (𝑓 “ ∪ 𝐷))) = ((𝑓 “ ∪ 𝐷) ∪ ran (◡𝑔 ↾ (𝐴 ∖ ∪ 𝐷)))) |
| 13 | 6, 12 | eqtr4id 2787 |
. 2
⊢ (((dom
𝑔 = 𝐵 ∧ ran 𝑔 ⊆ 𝐴) ∧ Fun ◡𝑔) → ran 𝐻 = ((𝑓 “ ∪ 𝐷) ∪ (𝐵 ∖ (𝑓 “ ∪ 𝐷)))) |
| 14 | | imassrn 6027 |
. . . 4
⊢ (𝑓 “ ∪ 𝐷)
⊆ ran 𝑓 |
| 15 | | sstr2 3937 |
. . . 4
⊢ ((𝑓 “ ∪ 𝐷)
⊆ ran 𝑓 → (ran
𝑓 ⊆ 𝐵 → (𝑓 “ ∪ 𝐷) ⊆ 𝐵)) |
| 16 | 14, 15 | ax-mp 5 |
. . 3
⊢ (ran
𝑓 ⊆ 𝐵 → (𝑓 “ ∪ 𝐷) ⊆ 𝐵) |
| 17 | | undif 4431 |
. . 3
⊢ ((𝑓 “ ∪ 𝐷)
⊆ 𝐵 ↔ ((𝑓 “ ∪ 𝐷)
∪ (𝐵 ∖ (𝑓 “ ∪ 𝐷)))
= 𝐵) |
| 18 | 16, 17 | sylib 218 |
. 2
⊢ (ran
𝑓 ⊆ 𝐵 → ((𝑓 “ ∪ 𝐷) ∪ (𝐵 ∖ (𝑓 “ ∪ 𝐷))) = 𝐵) |
| 19 | 13, 18 | sylan9eqr 2790 |
1
⊢ ((ran
𝑓 ⊆ 𝐵 ∧ ((dom 𝑔 = 𝐵 ∧ ran 𝑔 ⊆ 𝐴) ∧ Fun ◡𝑔)) → ran 𝐻 = 𝐵) |