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Metamath Proof Explorer

Theorem fdmeu

Description: There is exactly one codomain element for each element of the domain of a function. (Contributed by AV, 20-Apr-2025)

Ref Expression
Assertion fdmeu 
|- ( ( F : A --> B /\ X e. A ) -> E! y e. B ( F ` X ) = y )

Proof

Step Hyp Ref Expression
1 feu 
 |-  ( ( F : A --> B /\ X e. A ) -> E! y e. B <. X , y >. e. F )
2 ffn 
 |-  ( F : A --> B -> F Fn A )
3 2 anim1i 
 |-  ( ( F : A --> B /\ X e. A ) -> ( F Fn A /\ X e. A ) )
4 3 adantr 
 |-  ( ( ( F : A --> B /\ X e. A ) /\ y e. B ) -> ( F Fn A /\ X e. A ) )
5 fnopfvb 
 |-  ( ( F Fn A /\ X e. A ) -> ( ( F ` X ) = y <-> <. X , y >. e. F ) )
6 4 5 syl 
 |-  ( ( ( F : A --> B /\ X e. A ) /\ y e. B ) -> ( ( F ` X ) = y <-> <. X , y >. e. F ) )
7 6 reubidva 
 |-  ( ( F : A --> B /\ X e. A ) -> ( E! y e. B ( F ` X ) = y <-> E! y e. B <. X , y >. e. F ) )
8 1 7 mpbird 
 |-  ( ( F : A --> B /\ X e. A ) -> E! y e. B ( F ` X ) = y )