Metamath Proof Explorer


Theorem reldm

Description: An expression for the domain of a relation. (Contributed by NM, 22-Sep-2013) (Proof shortened by Sky Winter, 22-Oct-2026)

Ref Expression
Assertion reldm ( Rel 𝐴 → dom 𝐴 = ran ( 𝑥 ∈ 𝐴 ↦ ( 1st ‘ 𝑥 ) ) )

Proof

Step Hyp Ref Expression
1 releldm2 ⊢ ( Rel 𝐴 → ( 𝑦 ∈ dom 𝐴 ↔ ∃ 𝑥 ∈ 𝐴 ( 1st ‘ 𝑥 ) = 𝑦 ) )
2 eqid ⊢ ( 𝑥 ∈ 𝐴 ↦ ( 1st ‘ 𝑥 ) ) = ( 𝑥 ∈ 𝐴 ↦ ( 1st ‘ 𝑥 ) )
3 fvex ⊢ ( 1st ‘ 𝑥 ) ∈ V
4 2 3 elrnmpti ⊢ ( 𝑦 ∈ ran ( 𝑥 ∈ 𝐴 ↦ ( 1st ‘ 𝑥 ) ) ↔ ∃ 𝑥 ∈ 𝐴 𝑦 = ( 1st ‘ 𝑥 ) )
5 eqcom ⊢ ( 𝑦 = ( 1st ‘ 𝑥 ) ↔ ( 1st ‘ 𝑥 ) = 𝑦 )
6 5 rexbii ⊢ ( ∃ 𝑥 ∈ 𝐴 𝑦 = ( 1st ‘ 𝑥 ) ↔ ∃ 𝑥 ∈ 𝐴 ( 1st ‘ 𝑥 ) = 𝑦 )
7 4 6 bitri ⊢ ( 𝑦 ∈ ran ( 𝑥 ∈ 𝐴 ↦ ( 1st ‘ 𝑥 ) ) ↔ ∃ 𝑥 ∈ 𝐴 ( 1st ‘ 𝑥 ) = 𝑦 )
8 1 7 bitr4di ⊢ ( Rel 𝐴 → ( 𝑦 ∈ dom 𝐴 ↔ 𝑦 ∈ ran ( 𝑥 ∈ 𝐴 ↦ ( 1st ‘ 𝑥 ) ) ) )
9 8 eqrdv ⊢ ( Rel 𝐴 → dom 𝐴 = ran ( 𝑥 ∈ 𝐴 ↦ ( 1st ‘ 𝑥 ) ) )