Metamath Proof Explorer


Theorem bj-fvsnun1

Description: The value of a function with one of its ordered pairs replaced, at arguments other than the replaced one. (Contributed by NM, 23-Sep-2007) Put in deduction form and remove two sethood hypotheses. (Revised by BJ, 18-Mar-2023)

Ref Expression
Hypotheses bj-fvsnun.un
|- ( ph -> G = ( ( F |` ( C \ { A } ) ) u. { <. A , B >. } ) )
bj-fvsnun1.eldif
|- ( ph -> D e. ( C \ { A } ) )
Assertion bj-fvsnun1
|- ( ph -> ( G ` D ) = ( F ` D ) )

Proof

Step Hyp Ref Expression
1 bj-fvsnun.un
 |-  ( ph -> G = ( ( F |` ( C \ { A } ) ) u. { <. A , B >. } ) )
2 bj-fvsnun1.eldif
 |-  ( ph -> D e. ( C \ { A } ) )
3 eldifsnneq
 |-  ( D e. ( C \ { A } ) -> -. D = A )
4 2 3 syl
 |-  ( ph -> -. D = A )
5 1 4 bj-fununsn1
 |-  ( ph -> ( G ` D ) = ( ( F |` ( C \ { A } ) ) ` D ) )
6 2 fvresd
 |-  ( ph -> ( ( F |` ( C \ { A } ) ) ` D ) = ( F ` D ) )
7 5 6 eqtrd
 |-  ( ph -> ( G ` D ) = ( F ` D ) )