Metamath Proof Explorer


Theorem reldm

Description: An expression for the domain of a relation. (Contributed by NM, 22-Sep-2013)

Ref Expression
Assertion reldm ⊢ Rel ⁡ A → dom ⁡ A = ran ⁡ x ∈ A ⟼ 1 st ⁡ x

Proof

Step Hyp Ref Expression
1 releldm2 ⊢ Rel ⁡ A → y ∈ dom ⁡ A ↔ ∃ z ∈ A 1 st ⁡ z = y
2 fvex ⊢ 1 st ⁡ x ∈ V
3 eqid ⊢ x ∈ A ⟼ 1 st ⁡ x = x ∈ A ⟼ 1 st ⁡ x
4 2 3 fnmpti ⊢ x ∈ A ⟼ 1 st ⁡ x Fn A
5 fvelrnb ⊢ x ∈ A ⟼ 1 st ⁡ x Fn A → y ∈ ran ⁡ x ∈ A ⟼ 1 st ⁡ x ↔ ∃ z ∈ A x ∈ A ⟼ 1 st ⁡ x ⁡ z = y
6 4 5 ax-mp ⊢ y ∈ ran ⁡ x ∈ A ⟼ 1 st ⁡ x ↔ ∃ z ∈ A x ∈ A ⟼ 1 st ⁡ x ⁡ z = y
7 fveq2 ⊢ x = z → 1 st ⁡ x = 1 st ⁡ z
8 fvex ⊢ 1 st ⁡ z ∈ V
9 7 3 8 fvmpt ⊢ z ∈ A → x ∈ A ⟼ 1 st ⁡ x ⁡ z = 1 st ⁡ z
10 9 eqeq1d ⊢ z ∈ A → x ∈ A ⟼ 1 st ⁡ x ⁡ z = y ↔ 1 st ⁡ z = y
11 10 rexbiia ⊢ ∃ z ∈ A x ∈ A ⟼ 1 st ⁡ x ⁡ z = y ↔ ∃ z ∈ A 1 st ⁡ z = y
12 11 a1i ⊢ Rel ⁡ A → ∃ z ∈ A x ∈ A ⟼ 1 st ⁡ x ⁡ z = y ↔ ∃ z ∈ A 1 st ⁡ z = y
13 6 12 bitr2id ⊢ Rel ⁡ A → ∃ z ∈ A 1 st ⁡ z = y ↔ y ∈ ran ⁡ x ∈ A ⟼ 1 st ⁡ x
14 1 13 bitrd ⊢ Rel ⁡ A → y ∈ dom ⁡ A ↔ y ∈ ran ⁡ x ∈ A ⟼ 1 st ⁡ x
15 14 eqrdv ⊢ Rel ⁡ A → dom ⁡ A = ran ⁡ x ∈ A ⟼ 1 st ⁡ x