Metamath Proof Explorer


Theorem sbthlem4

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
Assertion sbthlem4 ⊢ dom ⁡ g = B ∧ ran ⁡ g ⊆ A ∧ Fun ⁡ g -1 → g -1 A ∖ ⋃ D = B ∖ f ⋃ D

Proof

Step Hyp Ref Expression
1 sbthlem.1 ⊢ A ∈ V
2 sbthlem.2 ⊢ D = x | x ⊆ A ∧ g B ∖ f x ⊆ A ∖ x
3 df-ima ⊢ g -1 A ∖ ⋃ D = ran ⁡ g -1 ↾ A ∖ ⋃ D
4 difss ⊢ B ∖ f ⋃ D ⊆ B
5 sseq2 ⊢ dom ⁡ g = B → B ∖ f ⋃ D ⊆ dom ⁡ g ↔ B ∖ f ⋃ D ⊆ B
6 4 5 mpbiri ⊢ dom ⁡ g = B → B ∖ f ⋃ D ⊆ dom ⁡ g
7 ssdmres ⊢ B ∖ f ⋃ D ⊆ dom ⁡ g ↔ dom ⁡ g ↾ B ∖ f ⋃ D = B ∖ f ⋃ D
8 6 7 sylib ⊢ dom ⁡ g = B → dom ⁡ g ↾ B ∖ f ⋃ D = B ∖ f ⋃ D
9 dfdm4 ⊢ dom ⁡ g ↾ B ∖ f ⋃ D = ran ⁡ g ↾ B ∖ f ⋃ D -1
10 8 9 eqtr3di ⊢ dom ⁡ g = B → B ∖ f ⋃ D = ran ⁡ g ↾ B ∖ f ⋃ D -1
11 funcnvres ⊢ Fun ⁡ g -1 → g ↾ B ∖ f ⋃ D -1 = g -1 ↾ g B ∖ f ⋃ D
12 1 2 sbthlem3 ⊢ ran ⁡ g ⊆ A → g B ∖ f ⋃ D = A ∖ ⋃ D
13 12 reseq2d ⊢ ran ⁡ g ⊆ A → g -1 ↾ g B ∖ f ⋃ D = g -1 ↾ A ∖ ⋃ D
14 11 13 sylan9eqr ⊢ ran ⁡ g ⊆ A ∧ Fun ⁡ g -1 → g ↾ B ∖ f ⋃ D -1 = g -1 ↾ A ∖ ⋃ D
15 14 rneqd ⊢ ran ⁡ g ⊆ A ∧ Fun ⁡ g -1 → ran ⁡ g ↾ B ∖ f ⋃ D -1 = ran ⁡ g -1 ↾ A ∖ ⋃ D
16 10 15 sylan9eq ⊢ dom ⁡ g = B ∧ ran ⁡ g ⊆ A ∧ Fun ⁡ g -1 → B ∖ f ⋃ D = ran ⁡ g -1 ↾ A ∖ ⋃ D
17 16 anassrs ⊢ dom ⁡ g = B ∧ ran ⁡ g ⊆ A ∧ Fun ⁡ g -1 → B ∖ f ⋃ D = ran ⁡ g -1 ↾ A ∖ ⋃ D
18 3 17 eqtr4id ⊢ dom ⁡ g = B ∧ ran ⁡ g ⊆ A ∧ Fun ⁡ g -1 → g -1 A ∖ ⋃ D = B ∖ f ⋃ D