reserve X,Y for set, p,x,x1,x2,y,y1,y2,z,z1,z2 for object;
reserve f,g,h for Function;

theorem
 g c= f & x in X & X /\ dom f c= dom g
  implies f.x = g.x
proof assume that
A1: g c= f and
A2: x in X and
A3: X /\ dom f c= dom g;
 per cases;
 suppose x in dom g;
 hence f.x = g.x by A1,Th2;
 end;
 suppose
A4:  not x in dom g;
  then not x in X /\ dom f by A3;
  then not x in dom f by A2,XBOOLE_0:def 4;
 hence f.x = {} by FUNCT_1:def 2
   .= g.x by A4,FUNCT_1:def 2;
 end;
end;
