
theorem Th28:
for X,Y be non empty set, x,y be set, E be Subset of [:X,Y:]
 holds chi(E,[:X,Y:]).(x,y) = chi(X-section(E,x),Y).y &
       chi(E,[:X,Y:]).(x,y) = chi(Y-section(E,y),X).x
proof
   let X,Y be non empty set, x,y be set, E be Subset of [:X,Y:];
   set z = [x,y];
   per cases;
   suppose A1: [x,y] in E; then
    consider x1,y1 be object such that
A2:  x1 in X & y1 in Y & [x,y] = [x1,y1] by ZFMISC_1:84;
    x = x1 & y = y1 by A2,XTUPLE_0:1; then
A3: y in X-section(E,x) & x in Y-section(E,y) by A1,A2;
    chi(E,[:X,Y:]).z = 1 by A1,RFUNCT_1:63;
    hence chi(E,[:X,Y:]).(x,y) = chi(X-section(E,x),Y).y &
          chi(E,[:X,Y:]).(x,y) = chi(Y-section(E,y),X).x by A3,RFUNCT_1:63;
   end;
   suppose A4: not [x,y] in E;
A5: chi(E,[:X,Y:]).(x,y) = 0
    proof
     per cases;
     suppose [x,y] in [:X,Y:];
      hence chi(E,[:X,Y:]).(x,y) = 0 by A4,FUNCT_3:def 3;
     end;
     suppose not [x,y] in [:X,Y:]; then
      not [x,y] in dom (chi(E,[:X,Y:]));
      hence chi(E,[:X,Y:]).(x,y) = 0 by FUNCT_1:def 2;
     end;
    end;
A6: now assume y in X-section(E,x); then
     ex y1 be Element of Y st y = y1 & [x,y1] in E;
     hence contradiction by A4;
    end;
A7: chi(X-section(E,x),Y).y = 0
    proof
     per cases;
     suppose y in Y;
      hence thesis by A6,FUNCT_3:def 3;
     end;
     suppose not y in Y; then
      not y in dom (chi(X-section(E,x),Y));
      hence thesis by FUNCT_1:def 2;
     end;
    end;
A8: now assume x in Y-section(E,y); then
     ex x1 be Element of X st x = x1 & [x1,y] in E;
     hence contradiction by A4;
    end;
    chi(Y-section(E,y),X).x = 0
    proof
     per cases;
     suppose x in X;
      hence thesis by A8,FUNCT_3:def 3;
     end;
     suppose not x in X; then
      not x in dom (chi(Y-section(E,y),X));
      hence thesis by FUNCT_1:def 2;
     end;
    end;
    hence chi(E,[:X,Y:]).(x,y) = chi(X-section(E,x),Y).y &
         chi(E,[:X,Y:]).(x,y) = chi(Y-section(E,y),X).x by A5,A7;
   end;
end;
