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Metamath Proof Explorer

Theorem mhm0

Description: A monoid homomorphism preserves zero. (Contributed by Mario Carneiro, 7-Mar-2015)

Ref Expression
Hypotheses mhm0.z 
|- .0. = ( 0g ` S )
mhm0.y 
|- Y = ( 0g ` T )
Assertion mhm0 
|- ( F e. ( S MndHom T ) -> ( F ` .0. ) = Y )

Proof

Step Hyp Ref Expression
1 mhm0.z 
 |-  .0. = ( 0g ` S )
2 mhm0.y 
 |-  Y = ( 0g ` T )
3 eqid 
 |-  ( Base ` S ) = ( Base ` S )
4 eqid 
 |-  ( Base ` T ) = ( Base ` T )
5 eqid 
 |-  ( +g ` S ) = ( +g ` S )
6 eqid 
 |-  ( +g ` T ) = ( +g ` T )
7 3 4 5 6 1 2 ismhm 
 |-  ( F e. ( S MndHom T ) <-> ( ( S e. Mnd /\ T e. Mnd ) /\ ( F : ( Base ` S ) --> ( Base ` T ) /\ A. x e. ( Base ` S ) A. y e. ( Base ` S ) ( F ` ( x ( +g ` S ) y ) ) = ( ( F ` x ) ( +g ` T ) ( F ` y ) ) /\ ( F ` .0. ) = Y ) ) )
8 7 simprbi 
 |-  ( F e. ( S MndHom T ) -> ( F : ( Base ` S ) --> ( Base ` T ) /\ A. x e. ( Base ` S ) A. y e. ( Base ` S ) ( F ` ( x ( +g ` S ) y ) ) = ( ( F ` x ) ( +g ` T ) ( F ` y ) ) /\ ( F ` .0. ) = Y ) )
9 8 simp3d 
 |-  ( F e. ( S MndHom T ) -> ( F ` .0. ) = Y )