Metamath Proof Explorer


Theorem cnvf1olem

Description: Lemma for cnvf1o . (Contributed by Mario Carneiro, 27-Apr-2014)

Ref Expression
Assertion cnvf1olem ⊢ Rel ⁡ A ∧ B ∈ A ∧ C = ⋃ B -1 → C ∈ A -1 ∧ B = ⋃ C -1

Proof

Step Hyp Ref Expression
1 simprr ⊢ Rel ⁡ A ∧ B ∈ A ∧ C = ⋃ B -1 → C = ⋃ B -1
2 1st2nd ⊢ Rel ⁡ A ∧ B ∈ A → B = 1 st ⁡ B 2 nd ⁡ B
3 2 adantrr ⊢ Rel ⁡ A ∧ B ∈ A ∧ C = ⋃ B -1 → B = 1 st ⁡ B 2 nd ⁡ B
4 3 sneqd ⊢ Rel ⁡ A ∧ B ∈ A ∧ C = ⋃ B -1 → B = 1 st ⁡ B 2 nd ⁡ B
5 4 cnveqd ⊢ Rel ⁡ A ∧ B ∈ A ∧ C = ⋃ B -1 → B -1 = 1 st ⁡ B 2 nd ⁡ B -1
6 5 unieqd ⊢ Rel ⁡ A ∧ B ∈ A ∧ C = ⋃ B -1 → ⋃ B -1 = ⋃ 1 st ⁡ B 2 nd ⁡ B -1
7 1 6 eqtrd ⊢ Rel ⁡ A ∧ B ∈ A ∧ C = ⋃ B -1 → C = ⋃ 1 st ⁡ B 2 nd ⁡ B -1
8 opswap ⊢ ⋃ 1 st ⁡ B 2 nd ⁡ B -1 = 2 nd ⁡ B 1 st ⁡ B
9 7 8 eqtrdi ⊢ Rel ⁡ A ∧ B ∈ A ∧ C = ⋃ B -1 → C = 2 nd ⁡ B 1 st ⁡ B
10 simprl ⊢ Rel ⁡ A ∧ B ∈ A ∧ C = ⋃ B -1 → B ∈ A
11 3 10 eqeltrrd ⊢ Rel ⁡ A ∧ B ∈ A ∧ C = ⋃ B -1 → 1 st ⁡ B 2 nd ⁡ B ∈ A
12 fvex ⊢ 2 nd ⁡ B ∈ V
13 fvex ⊢ 1 st ⁡ B ∈ V
14 12 13 opelcnv ⊢ 2 nd ⁡ B 1 st ⁡ B ∈ A -1 ↔ 1 st ⁡ B 2 nd ⁡ B ∈ A
15 11 14 sylibr ⊢ Rel ⁡ A ∧ B ∈ A ∧ C = ⋃ B -1 → 2 nd ⁡ B 1 st ⁡ B ∈ A -1
16 9 15 eqeltrd ⊢ Rel ⁡ A ∧ B ∈ A ∧ C = ⋃ B -1 → C ∈ A -1
17 opswap ⊢ ⋃ 2 nd ⁡ B 1 st ⁡ B -1 = 1 st ⁡ B 2 nd ⁡ B
18 17 eqcomi ⊢ 1 st ⁡ B 2 nd ⁡ B = ⋃ 2 nd ⁡ B 1 st ⁡ B -1
19 9 sneqd ⊢ Rel ⁡ A ∧ B ∈ A ∧ C = ⋃ B -1 → C = 2 nd ⁡ B 1 st ⁡ B
20 19 cnveqd ⊢ Rel ⁡ A ∧ B ∈ A ∧ C = ⋃ B -1 → C -1 = 2 nd ⁡ B 1 st ⁡ B -1
21 20 unieqd ⊢ Rel ⁡ A ∧ B ∈ A ∧ C = ⋃ B -1 → ⋃ C -1 = ⋃ 2 nd ⁡ B 1 st ⁡ B -1
22 18 3 21 3eqtr4a ⊢ Rel ⁡ A ∧ B ∈ A ∧ C = ⋃ B -1 → B = ⋃ C -1
23 16 22 jca ⊢ Rel ⁡ A ∧ B ∈ A ∧ C = ⋃ B -1 → C ∈ A -1 ∧ B = ⋃ C -1