Metamath Proof Explorer


Theorem foelrnf

Description: Property of a surjective function. As foelrn but with less disjoint vars constraints. (Contributed by Glauco Siliprandi, 17-Aug-2020)

Ref Expression
Hypothesis foelrnf.1 𝑥 𝐹
Assertion foelrnf ( ( 𝐹 : 𝐴onto𝐵𝐶𝐵 ) → ∃ 𝑥𝐴 𝐶 = ( 𝐹𝑥 ) )

Proof

Step Hyp Ref Expression
1 foelrnf.1 𝑥 𝐹
2 1 dffo3f ( 𝐹 : 𝐴onto𝐵 ↔ ( 𝐹 : 𝐴𝐵 ∧ ∀ 𝑦𝐵𝑥𝐴 𝑦 = ( 𝐹𝑥 ) ) )
3 2 simprbi ( 𝐹 : 𝐴onto𝐵 → ∀ 𝑦𝐵𝑥𝐴 𝑦 = ( 𝐹𝑥 ) )
4 eqeq1 ( 𝑦 = 𝐶 → ( 𝑦 = ( 𝐹𝑥 ) ↔ 𝐶 = ( 𝐹𝑥 ) ) )
5 4 rexbidv ( 𝑦 = 𝐶 → ( ∃ 𝑥𝐴 𝑦 = ( 𝐹𝑥 ) ↔ ∃ 𝑥𝐴 𝐶 = ( 𝐹𝑥 ) ) )
6 5 rspccva ( ( ∀ 𝑦𝐵𝑥𝐴 𝑦 = ( 𝐹𝑥 ) ∧ 𝐶𝐵 ) → ∃ 𝑥𝐴 𝐶 = ( 𝐹𝑥 ) )
7 3 6 sylan ( ( 𝐹 : 𝐴onto𝐵𝐶𝐵 ) → ∃ 𝑥𝐴 𝐶 = ( 𝐹𝑥 ) )