Metamath Proof Explorer


Theorem reuf1od

Description: There is exactly one element in each of two isomorphic sets. (Contributed by AV, 19-Mar-2023)

Ref Expression
Hypotheses reuf1od.f ⊢ φ → F : C ⟶ 1-1 onto B
reuf1od.x ⊢ φ ∧ x = F ⁡ y → ψ ↔ χ
Assertion reuf1od ⊢ φ → ∃! x ∈ B ψ ↔ ∃! y ∈ C χ

Proof

Step Hyp Ref Expression
1 reuf1od.f ⊢ φ → F : C ⟶ 1-1 onto B
2 reuf1od.x ⊢ φ ∧ x = F ⁡ y → ψ ↔ χ
3 f1of ⊢ F : C ⟶ 1-1 onto B → F : C ⟶ B
4 1 3 syl ⊢ φ → F : C ⟶ B
5 4 ffvelcdmda ⊢ φ ∧ y ∈ C → F ⁡ y ∈ B
6 f1ofveu ⊢ F : C ⟶ 1-1 onto B ∧ x ∈ B → ∃! y ∈ C F ⁡ y = x
7 eqcom ⊢ x = F ⁡ y ↔ F ⁡ y = x
8 7 reubii ⊢ ∃! y ∈ C x = F ⁡ y ↔ ∃! y ∈ C F ⁡ y = x
9 6 8 sylibr ⊢ F : C ⟶ 1-1 onto B ∧ x ∈ B → ∃! y ∈ C x = F ⁡ y
10 1 9 sylan ⊢ φ ∧ x ∈ B → ∃! y ∈ C x = F ⁡ y
11 5 10 2 reuxfr1d ⊢ φ → ∃! x ∈ B ψ ↔ ∃! y ∈ C χ