Metamath Proof Explorer


Theorem sbthlem6

Description: Lemma for sbth . (Contributed by NM, 27-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 sbthlem6 ⊢ ran ⁡ f ⊆ B ∧ dom ⁡ g = B ∧ ran ⁡ g ⊆ A ∧ Fun ⁡ g -1 → ran ⁡ H = B

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 rnun ⊢ ran ⁡ f ↾ ⋃ D ∪ g -1 ↾ A ∖ ⋃ D = ran ⁡ f ↾ ⋃ D ∪ ran ⁡ g -1 ↾ A ∖ ⋃ D
5 3 rneqi ⊢ ran ⁡ H = ran ⁡ f ↾ ⋃ D ∪ g -1 ↾ A ∖ ⋃ D
6 df-ima ⊢ f ⋃ D = ran ⁡ f ↾ ⋃ D
7 6 uneq1i ⊢ f ⋃ D ∪ ran ⁡ g -1 ↾ A ∖ ⋃ D = ran ⁡ f ↾ ⋃ D ∪ ran ⁡ g -1 ↾ A ∖ ⋃ D
8 4 5 7 3eqtr4i ⊢ ran ⁡ H = f ⋃ D ∪ ran ⁡ g -1 ↾ A ∖ ⋃ D
9 1 2 sbthlem4 ⊢ dom ⁡ g = B ∧ ran ⁡ g ⊆ A ∧ Fun ⁡ g -1 → g -1 A ∖ ⋃ D = B ∖ f ⋃ D
10 df-ima ⊢ g -1 A ∖ ⋃ D = ran ⁡ g -1 ↾ A ∖ ⋃ D
11 9 10 eqtr3di ⊢ dom ⁡ g = B ∧ ran ⁡ g ⊆ A ∧ Fun ⁡ g -1 → B ∖ f ⋃ D = ran ⁡ g -1 ↾ A ∖ ⋃ D
12 11 uneq2d ⊢ dom ⁡ g = B ∧ ran ⁡ g ⊆ A ∧ Fun ⁡ g -1 → f ⋃ D ∪ B ∖ f ⋃ D = f ⋃ D ∪ ran ⁡ g -1 ↾ A ∖ ⋃ D
13 8 12 eqtr4id ⊢ dom ⁡ g = B ∧ ran ⁡ g ⊆ A ∧ Fun ⁡ g -1 → ran ⁡ H = f ⋃ D ∪ B ∖ f ⋃ D
14 imassrn ⊢ f ⋃ D ⊆ ran ⁡ f
15 sstr2 ⊢ f ⋃ D ⊆ ran ⁡ f → ran ⁡ f ⊆ B → f ⋃ D ⊆ B
16 14 15 ax-mp ⊢ ran ⁡ f ⊆ B → f ⋃ D ⊆ B
17 undif ⊢ f ⋃ D ⊆ B ↔ f ⋃ D ∪ B ∖ f ⋃ D = B
18 16 17 sylib ⊢ ran ⁡ f ⊆ B → f ⋃ D ∪ B ∖ f ⋃ D = B
19 13 18 sylan9eqr ⊢ ran ⁡ f ⊆ B ∧ dom ⁡ g = B ∧ ran ⁡ g ⊆ A ∧ Fun ⁡ g -1 → ran ⁡ H = B