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Metamath Proof Explorer

Theorem mhmf

Description: A monoid homomorphism is a function. (Contributed by Mario Carneiro, 7-Mar-2015)

Ref Expression
Hypotheses mhmf.b 
|- B = ( Base ` S )
mhmf.c 
|- C = ( Base ` T )
Assertion mhmf 
|- ( F e. ( S MndHom T ) -> F : B --> C )

Proof

Step Hyp Ref Expression
1 mhmf.b 
 |-  B = ( Base ` S )
2 mhmf.c 
 |-  C = ( Base ` T )
3 eqid 
 |-  ( +g ` S ) = ( +g ` S )
4 eqid 
 |-  ( +g ` T ) = ( +g ` T )
5 eqid 
 |-  ( 0g ` S ) = ( 0g ` S )
6 eqid 
 |-  ( 0g ` T ) = ( 0g ` T )
7 1 2 3 4 5 6 ismhm 
 |-  ( F e. ( S MndHom T ) <-> ( ( S e. Mnd /\ T e. Mnd ) /\ ( F : B --> C /\ A. x e. B A. y e. B ( F ` ( x ( +g ` S ) y ) ) = ( ( F ` x ) ( +g ` T ) ( F ` y ) ) /\ ( F ` ( 0g ` S ) ) = ( 0g ` T ) ) ) )
8 7 simprbi 
 |-  ( F e. ( S MndHom T ) -> ( F : B --> C /\ A. x e. B A. y e. B ( F ` ( x ( +g ` S ) y ) ) = ( ( F ` x ) ( +g ` T ) ( F ` y ) ) /\ ( F ` ( 0g ` S ) ) = ( 0g ` T ) ) )
9 8 simp1d 
 |-  ( F e. ( S MndHom T ) -> F : B --> C )