let C be non empty set ; :: thesis: for f, g, h being Membership_Func of C holds (f * g) * h = f * (g * h)
let f, g, h be Membership_Func of C; :: thesis: (f * g) * h = f * (g * h)
A1: ( C = dom ((f * g) * h) & C = dom (f * (g * h)) ) by FUNCT_2:def 1;
for c being Element of C st c in C holds
((f * g) * h) . c = (f * (g * h)) . c

proof

let c be Element of C; :: thesis: ( c in C implies ((f * g) * h) . c = (f * (g * h)) . c )
((f * g) * h) . c = ((f * g) . c) * (h . c) by Def2
.= ((f . c) * (g . c)) * (h . c) by Def2
.= (f . c) * ((g . c) * (h . c))
.= (f . c) * ((g * h) . c) by Def2 ;
hence ( c in C implies ((f * g) * h) . c = (f * (g * h)) . c ) by Def2; :: thesis: verum

end;

hence (f * g) * h = f * (g * h) by A1, PARTFUN1:34; :: thesis: verum