Metamath Proof Explorer


Theorem eldmrexrn

Description: For any element in the domain of a function there is an element in the range of the function which is the function value for the element of the domain. (Contributed by Alexander van der Vekens, 8-Dec-2017)

Ref Expression
Assertion eldmrexrn ( Fun 𝐹 → ( 𝑌 ∈ dom 𝐹 → ∃ 𝑥 ∈ ran 𝐹 𝑥 = ( 𝐹 ‘ 𝑌 ) ) )

Proof

Step Hyp Ref Expression
1 fvelrn ⊢ ( ( Fun 𝐹 ∧ 𝑌 ∈ dom 𝐹 ) → ( 𝐹 ‘ 𝑌 ) ∈ ran 𝐹 )
2 eqid ⊢ ( 𝐹 ‘ 𝑌 ) = ( 𝐹 ‘ 𝑌 )
3 eqeq1 ⊢ ( 𝑥 = ( 𝐹 ‘ 𝑌 ) → ( 𝑥 = ( 𝐹 ‘ 𝑌 ) ↔ ( 𝐹 ‘ 𝑌 ) = ( 𝐹 ‘ 𝑌 ) ) )
4 3 rspcev ⊢ ( ( ( 𝐹 ‘ 𝑌 ) ∈ ran 𝐹 ∧ ( 𝐹 ‘ 𝑌 ) = ( 𝐹 ‘ 𝑌 ) ) → ∃ 𝑥 ∈ ran 𝐹 𝑥 = ( 𝐹 ‘ 𝑌 ) )
5 1 2 4 sylancl ⊢ ( ( Fun 𝐹 ∧ 𝑌 ∈ dom 𝐹 ) → ∃ 𝑥 ∈ ran 𝐹 𝑥 = ( 𝐹 ‘ 𝑌 ) )
6 5 ex ⊢ ( Fun 𝐹 → ( 𝑌 ∈ dom 𝐹 → ∃ 𝑥 ∈ ran 𝐹 𝑥 = ( 𝐹 ‘ 𝑌 ) ) )