reserve p,q,x,x1,x2,y,y1,y2,z,z1,z2 for set;
reserve A,B,V,X,X1,X2,Y,Y1,Y2,Z for set;
reserve C,C1,C2,D,D1,D2 for non empty set;
reserve A1,A2,B1,B2 for non empty set,
  f for Function of A1,B1,
  g for Function of A2,B2,
  Y1 for non empty Subset of A1,
  Y2 for non empty Subset of A2;

theorem
  [:f,g:]|([:Y1,Y2:] qua Subset of [:A1,A2:])
qua Function of [:Y1,Y2:],[:B1,B2:]
= [:f|Y1,g|Y2:] qua Function of [:Y1,Y2:],[:B1,B2:]
proof
 let a be Element of [:Y1,Y2:];
  consider a1 being Element of Y1, a2 being Element of Y2 such that
A1: a = [a1,a2] by DOMAIN_1:1;
A2: (g|Y2).a2 = g.a2 by FUNCT_1:49;
A3: (f|Y1).a1 = f.a1 by FUNCT_1:49;
 thus [:f|Y1,g|Y2:].a = [:f|Y1,g|Y2:].(a1,a2) by A1
      .= [f.a1,g.a2] by A3,A2,Th75
      .= [:f,g:].(a1,a2) by Th75
      .= ([:f,g:]|[:Y1,Y2:]).a by A1,FUNCT_1:49;
end;
