let X be set ; :: thesis: for D, C being non empty set
for g, f being PartFunc of C,D holds
( g = f | X iff ( dom g = (dom f) /\ X & ( for c being Element of C st c in dom g holds
g /. c = f /. c ) ) )
let D, C be non empty set ; :: thesis: for g, f being PartFunc of C,D holds
( g = f | X iff ( dom g = (dom f) /\ X & ( for c being Element of C st c in dom g holds
g /. c = f /. c ) ) )
let g, f be PartFunc of C,D; :: thesis: ( g = f | X iff ( dom g = (dom f) /\ X & ( for c being Element of C st c in dom g holds
g /. c = f /. c ) ) )
thus ( g = f | X implies ( dom g = (dom f) /\ X & ( for c being Element of C st c in dom g holds
g /. c = f /. c ) ) ) :: thesis: ( dom g = (dom f) /\ X & ( for c being Element of C st c in dom g holds
g /. c = f /. c ) implies g = f | X )

proof

assume A1: g = f | X ; :: thesis: ( dom g = (dom f) /\ X & ( for c being Element of C st c in dom g holds
g /. c = f /. c ) )
hence dom g = (dom f) /\ X by RELAT_1:90; :: thesis: for c being Element of C st c in dom g holds
g /. c = f /. c
let c be Element of C; :: thesis: ( c in dom g implies g /. c = f /. c )
assume A2: c in dom g ; :: thesis: g /. c = f /. c
then g . c = f . c by A1, FUNCT_1:70;
then A3: g /. c = f . c by A2, PARTFUN1:def 8;
dom g = (dom f) /\ X by A1, RELAT_1:90;
then c in dom f by A2, XBOOLE_0:def 4;
hence g /. c = f /. c by A3, PARTFUN1:def 8; :: thesis: verum

end;

assume that
A4: dom g = (dom f) /\ X and
A5: for c being Element of C st c in dom g holds
g /. c = f /. c ; :: thesis: g = f | X

now

let x be set ; :: thesis: ( x in dom g implies g . x = f . x )
assume A6: x in dom g ; :: thesis: g . x = f . x
then reconsider y = x as Element of C ;
g /. y = f /. y by A5, A6;
then A7: g . y = f /. y by A6, PARTFUN1:def 8;
x in dom f by A4, A6, XBOOLE_0:def 4;
hence g . x = f . x by A7, PARTFUN1:def 8; :: thesis: verum

end;

hence g = f | X by A4, FUNCT_1:68; :: thesis: verum