| Step | Hyp | Ref
 | Expression | 
| 1 |   | sbthlem.1 | 
. . . . . . 7
⊢ 𝐴 ∈ V | 
| 2 |   | sbthlem.2 | 
. . . . . . 7
⊢ 𝐷 = {𝑥 ∣ (𝑥 ⊆ 𝐴 ∧ (𝑔 “ (𝐵 ∖ (𝑓 “ 𝑥))) ⊆ (𝐴 ∖ 𝑥))} | 
| 3 | 1, 2 | sbthlem2 7024 | 
. . . . . 6
⊢ (ran
𝑔 ⊆ 𝐴 → (𝐴 ∖ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷)))) ⊆ ∪ 𝐷) | 
| 4 | 1, 2 | sbthlem1 7023 | 
. . . . . 6
⊢ ∪ 𝐷
⊆ (𝐴 ∖ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷)))) | 
| 5 | 3, 4 | jctil 312 | 
. . . . 5
⊢ (ran
𝑔 ⊆ 𝐴 → (∪ 𝐷 ⊆ (𝐴 ∖ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷)))) ∧ (𝐴 ∖ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷)))) ⊆ ∪ 𝐷)) | 
| 6 |   | eqss 3198 | 
. . . . 5
⊢ (∪ 𝐷 =
(𝐴 ∖ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷)))) ↔ (∪ 𝐷
⊆ (𝐴 ∖ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷)))) ∧ (𝐴 ∖ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷)))) ⊆ ∪ 𝐷)) | 
| 7 | 5, 6 | sylibr 134 | 
. . . 4
⊢ (ran
𝑔 ⊆ 𝐴 → ∪ 𝐷 = (𝐴 ∖ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷))))) | 
| 8 | 7 | difeq2d 3281 | 
. . 3
⊢ (ran
𝑔 ⊆ 𝐴 → (𝐴 ∖ ∪ 𝐷) = (𝐴 ∖ (𝐴 ∖ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷)))))) | 
| 9 | 8 | adantl 277 | 
. 2
⊢
((EXMID ∧ ran 𝑔 ⊆ 𝐴) → (𝐴 ∖ ∪ 𝐷) = (𝐴 ∖ (𝐴 ∖ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷)))))) | 
| 10 |   | imassrn 5020 | 
. . . . 5
⊢ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷))) ⊆ ran 𝑔 | 
| 11 |   | sstr2 3190 | 
. . . . 5
⊢ ((𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷))) ⊆ ran 𝑔 → (ran 𝑔 ⊆ 𝐴 → (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷))) ⊆ 𝐴)) | 
| 12 | 10, 11 | ax-mp 5 | 
. . . 4
⊢ (ran
𝑔 ⊆ 𝐴 → (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷))) ⊆ 𝐴) | 
| 13 |   | exmidexmid 4229 | 
. . . . . . 7
⊢
(EXMID → DECID 𝑦 ∈ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷)))) | 
| 14 |   | dcstab 845 | 
. . . . . . 7
⊢
(DECID 𝑦 ∈ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷))) → STAB
𝑦 ∈ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷)))) | 
| 15 | 13, 14 | syl 14 | 
. . . . . 6
⊢
(EXMID → STAB 𝑦 ∈ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷)))) | 
| 16 | 15 | alrimiv 1888 | 
. . . . 5
⊢
(EXMID → ∀𝑦STAB 𝑦 ∈ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷)))) | 
| 17 |   | dfss4st 3396 | 
. . . . 5
⊢
(∀𝑦STAB 𝑦 ∈ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷))) → ((𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷))) ⊆ 𝐴 ↔ (𝐴 ∖ (𝐴 ∖ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷))))) = (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷))))) | 
| 18 | 16, 17 | syl 14 | 
. . . 4
⊢
(EXMID → ((𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷))) ⊆ 𝐴 ↔ (𝐴 ∖ (𝐴 ∖ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷))))) = (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷))))) | 
| 19 | 12, 18 | imbitrid 154 | 
. . 3
⊢
(EXMID → (ran 𝑔 ⊆ 𝐴 → (𝐴 ∖ (𝐴 ∖ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷))))) = (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷))))) | 
| 20 | 19 | imp 124 | 
. 2
⊢
((EXMID ∧ ran 𝑔 ⊆ 𝐴) → (𝐴 ∖ (𝐴 ∖ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷))))) = (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷)))) | 
| 21 | 9, 20 | eqtr2d 2230 | 
1
⊢
((EXMID ∧ ran 𝑔 ⊆ 𝐴) → (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷))) = (𝐴 ∖ ∪ 𝐷)) |