Metamath Proof Explorer


Theorem elrnrexdm

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

Ref Expression
Assertion elrnrexdm ( Fun 𝐹 → ( 𝑌 ∈ ran 𝐹 → ∃ 𝑥 ∈ dom 𝐹 𝑌 = ( 𝐹𝑥 ) ) )

Proof

Step Hyp Ref Expression
1 eqidd ( 𝑌 ∈ ran 𝐹𝑌 = 𝑌 )
2 1 ancli ( 𝑌 ∈ ran 𝐹 → ( 𝑌 ∈ ran 𝐹𝑌 = 𝑌 ) )
3 2 adantl ( ( Fun 𝐹𝑌 ∈ ran 𝐹 ) → ( 𝑌 ∈ ran 𝐹𝑌 = 𝑌 ) )
4 eqeq2 ( 𝑦 = 𝑌 → ( 𝑌 = 𝑦𝑌 = 𝑌 ) )
5 4 rspcev ( ( 𝑌 ∈ ran 𝐹𝑌 = 𝑌 ) → ∃ 𝑦 ∈ ran 𝐹 𝑌 = 𝑦 )
6 3 5 syl ( ( Fun 𝐹𝑌 ∈ ran 𝐹 ) → ∃ 𝑦 ∈ ran 𝐹 𝑌 = 𝑦 )
7 6 ex ( Fun 𝐹 → ( 𝑌 ∈ ran 𝐹 → ∃ 𝑦 ∈ ran 𝐹 𝑌 = 𝑦 ) )
8 funfn ( Fun 𝐹𝐹 Fn dom 𝐹 )
9 eqeq2 ( 𝑦 = ( 𝐹𝑥 ) → ( 𝑌 = 𝑦𝑌 = ( 𝐹𝑥 ) ) )
10 9 rexrn ( 𝐹 Fn dom 𝐹 → ( ∃ 𝑦 ∈ ran 𝐹 𝑌 = 𝑦 ↔ ∃ 𝑥 ∈ dom 𝐹 𝑌 = ( 𝐹𝑥 ) ) )
11 8 10 sylbi ( Fun 𝐹 → ( ∃ 𝑦 ∈ ran 𝐹 𝑌 = 𝑦 ↔ ∃ 𝑥 ∈ dom 𝐹 𝑌 = ( 𝐹𝑥 ) ) )
12 7 11 sylibd ( Fun 𝐹 → ( 𝑌 ∈ ran 𝐹 → ∃ 𝑥 ∈ dom 𝐹 𝑌 = ( 𝐹𝑥 ) ) )