Metamath Proof Explorer


Theorem f1oresf1o

Description: Build a bijection by restricting the domain of a bijection. (Contributed by AV, 31-Jul-2022)

Ref Expression
Hypotheses f1oresf1o.1 ⊢ φ → F : A ⟶ 1-1 onto B
f1oresf1o.2 ⊢ φ → D ⊆ A
f1oresf1o.3 ⊢ φ → ∃ x ∈ D F ⁡ x = y ↔ y ∈ B ∧ χ
Assertion f1oresf1o ⊢ φ → F ↾ D : D ⟶ 1-1 onto y ∈ B | χ

Proof

Step Hyp Ref Expression
1 f1oresf1o.1 ⊢ φ → F : A ⟶ 1-1 onto B
2 f1oresf1o.2 ⊢ φ → D ⊆ A
3 f1oresf1o.3 ⊢ φ → ∃ x ∈ D F ⁡ x = y ↔ y ∈ B ∧ χ
4 f1of1 ⊢ F : A ⟶ 1-1 onto B → F : A ⟶ 1-1 B
5 1 4 syl ⊢ φ → F : A ⟶ 1-1 B
6 f1ores ⊢ F : A ⟶ 1-1 B ∧ D ⊆ A → F ↾ D : D ⟶ 1-1 onto F D
7 5 2 6 syl2anc ⊢ φ → F ↾ D : D ⟶ 1-1 onto F D
8 f1ofun ⊢ F : A ⟶ 1-1 onto B → Fun ⁡ F
9 1 8 syl ⊢ φ → Fun ⁡ F
10 f1odm ⊢ F : A ⟶ 1-1 onto B → dom ⁡ F = A
11 1 10 syl ⊢ φ → dom ⁡ F = A
12 2 11 sseqtrrd ⊢ φ → D ⊆ dom ⁡ F
13 dfimafn ⊢ Fun ⁡ F ∧ D ⊆ dom ⁡ F → F D = y | ∃ x ∈ D F ⁡ x = y
14 9 12 13 syl2anc ⊢ φ → F D = y | ∃ x ∈ D F ⁡ x = y
15 3 abbidv ⊢ φ → y | ∃ x ∈ D F ⁡ x = y = y | y ∈ B ∧ χ
16 df-rab ⊢ y ∈ B | χ = y | y ∈ B ∧ χ
17 15 16 eqtr4di ⊢ φ → y | ∃ x ∈ D F ⁡ x = y = y ∈ B | χ
18 14 17 eqtr2d ⊢ φ → y ∈ B | χ = F D
19 18 f1oeq3d ⊢ φ → F ↾ D : D ⟶ 1-1 onto y ∈ B | χ ↔ F ↾ D : D ⟶ 1-1 onto F D
20 7 19 mpbird ⊢ φ → F ↾ D : D ⟶ 1-1 onto y ∈ B | χ