Metamath Proof Explorer


Theorem sbthlem5

Description: Lemma for sbth . (Contributed by NM, 22-Mar-1998)

Ref Expression
Hypotheses sbthlem.1 ⊢ A ∈ V
sbthlem.2 ⊢ D = x | x ⊆ A ∧ g B ∖ f x ⊆ A ∖ x
sbthlem.3 ⊢ H = f ↾ ⋃ D ∪ g -1 ↾ A ∖ ⋃ D
Assertion sbthlem5 ⊢ dom ⁡ f = A ∧ ran ⁡ g ⊆ A → dom ⁡ H = A

Proof

Step Hyp Ref Expression
1 sbthlem.1 ⊢ A ∈ V
2 sbthlem.2 ⊢ D = x | x ⊆ A ∧ g B ∖ f x ⊆ A ∖ x
3 sbthlem.3 ⊢ H = f ↾ ⋃ D ∪ g -1 ↾ A ∖ ⋃ D
4 3 dmeqi ⊢ dom ⁡ H = dom ⁡ f ↾ ⋃ D ∪ g -1 ↾ A ∖ ⋃ D
5 dmun ⊢ dom ⁡ f ↾ ⋃ D ∪ g -1 ↾ A ∖ ⋃ D = dom ⁡ f ↾ ⋃ D ∪ dom ⁡ g -1 ↾ A ∖ ⋃ D
6 dmres ⊢ dom ⁡ f ↾ ⋃ D = ⋃ D ∩ dom ⁡ f
7 dmres ⊢ dom ⁡ g -1 ↾ A ∖ ⋃ D = A ∖ ⋃ D ∩ dom ⁡ g -1
8 df-rn ⊢ ran ⁡ g = dom ⁡ g -1
9 8 eqcomi ⊢ dom ⁡ g -1 = ran ⁡ g
10 9 ineq2i ⊢ A ∖ ⋃ D ∩ dom ⁡ g -1 = A ∖ ⋃ D ∩ ran ⁡ g
11 7 10 eqtri ⊢ dom ⁡ g -1 ↾ A ∖ ⋃ D = A ∖ ⋃ D ∩ ran ⁡ g
12 6 11 uneq12i ⊢ dom ⁡ f ↾ ⋃ D ∪ dom ⁡ g -1 ↾ A ∖ ⋃ D = ⋃ D ∩ dom ⁡ f ∪ A ∖ ⋃ D ∩ ran ⁡ g
13 5 12 eqtri ⊢ dom ⁡ f ↾ ⋃ D ∪ g -1 ↾ A ∖ ⋃ D = ⋃ D ∩ dom ⁡ f ∪ A ∖ ⋃ D ∩ ran ⁡ g
14 4 13 eqtri ⊢ dom ⁡ H = ⋃ D ∩ dom ⁡ f ∪ A ∖ ⋃ D ∩ ran ⁡ g
15 1 2 sbthlem1 ⊢ ⋃ D ⊆ A ∖ g B ∖ f ⋃ D
16 difss ⊢ A ∖ g B ∖ f ⋃ D ⊆ A
17 15 16 sstri ⊢ ⋃ D ⊆ A
18 sseq2 ⊢ dom ⁡ f = A → ⋃ D ⊆ dom ⁡ f ↔ ⋃ D ⊆ A
19 17 18 mpbiri ⊢ dom ⁡ f = A → ⋃ D ⊆ dom ⁡ f
20 dfss ⊢ ⋃ D ⊆ dom ⁡ f ↔ ⋃ D = ⋃ D ∩ dom ⁡ f
21 19 20 sylib ⊢ dom ⁡ f = A → ⋃ D = ⋃ D ∩ dom ⁡ f
22 21 uneq1d ⊢ dom ⁡ f = A → ⋃ D ∪ A ∖ ⋃ D = ⋃ D ∩ dom ⁡ f ∪ A ∖ ⋃ D
23 1 2 sbthlem3 ⊢ ran ⁡ g ⊆ A → g B ∖ f ⋃ D = A ∖ ⋃ D
24 imassrn ⊢ g B ∖ f ⋃ D ⊆ ran ⁡ g
25 23 24 eqsstrrdi ⊢ ran ⁡ g ⊆ A → A ∖ ⋃ D ⊆ ran ⁡ g
26 dfss ⊢ A ∖ ⋃ D ⊆ ran ⁡ g ↔ A ∖ ⋃ D = A ∖ ⋃ D ∩ ran ⁡ g
27 25 26 sylib ⊢ ran ⁡ g ⊆ A → A ∖ ⋃ D = A ∖ ⋃ D ∩ ran ⁡ g
28 27 uneq2d ⊢ ran ⁡ g ⊆ A → ⋃ D ∩ dom ⁡ f ∪ A ∖ ⋃ D = ⋃ D ∩ dom ⁡ f ∪ A ∖ ⋃ D ∩ ran ⁡ g
29 22 28 sylan9eq ⊢ dom ⁡ f = A ∧ ran ⁡ g ⊆ A → ⋃ D ∪ A ∖ ⋃ D = ⋃ D ∩ dom ⁡ f ∪ A ∖ ⋃ D ∩ ran ⁡ g
30 14 29 eqtr4id ⊢ dom ⁡ f = A ∧ ran ⁡ g ⊆ A → dom ⁡ H = ⋃ D ∪ A ∖ ⋃ D
31 undif ⊢ ⋃ D ⊆ A ↔ ⋃ D ∪ A ∖ ⋃ D = A
32 17 31 mpbi ⊢ ⋃ D ∪ A ∖ ⋃ D = A
33 30 32 eqtrdi ⊢ dom ⁡ f = A ∧ ran ⁡ g ⊆ A → dom ⁡ H = A