Metamath Proof Explorer


Theorem fconstmpt

Description: Representation of a constant function using the mapping operation. (Note that x cannot appear free in B .) (Contributed by NM, 12-Oct-1999) (Revised by Mario Carneiro, 16-Nov-2013)

Ref Expression
Assertion fconstmpt
|- ( A X. { B } ) = ( x e. A |-> B )

Proof

Step Hyp Ref Expression
1 velsn
 |-  ( y e. { B } <-> y = B )
2 1 anbi2i
 |-  ( ( x e. A /\ y e. { B } ) <-> ( x e. A /\ y = B ) )
3 2 opabbii
 |-  { <. x , y >. | ( x e. A /\ y e. { B } ) } = { <. x , y >. | ( x e. A /\ y = B ) }
4 df-xp
 |-  ( A X. { B } ) = { <. x , y >. | ( x e. A /\ y e. { B } ) }
5 df-mpt
 |-  ( x e. A |-> B ) = { <. x , y >. | ( x e. A /\ y = B ) }
6 3 4 5 3eqtr4i
 |-  ( A X. { B } ) = ( x e. A |-> B )