
theorem Th19:
for X,Y be non empty set, E be Subset of [:X,Y:], p be set holds
  ( p in X implies X-section([:X,Y:] \ E,p) = Y \ X-section(E,p) )
& ( p in Y implies Y-section([:X,Y:] \ E,p) = X \ Y-section(E,p) )
proof
   let X,Y be non empty set, E be Subset of [:X,Y:], p be set;
   hereby assume A1: p in X;
    now let y be set;
     assume A2: y in X-section([:X,Y:] \ E,p); then
A3:   ex y1 be Element of Y st y = y1 & [p,y1] in [:X,Y:] \ E;
     now assume y in X-section(E,p); then
      ex y2 be Element of Y st y = y2 & [p,y2] in E;
      hence contradiction by A3,XBOOLE_0:def 5;
     end;
     hence y in Y \ X-section(E,p) by A2,XBOOLE_0:def 5;
    end; then
A4: X-section([:X,Y:] \ E,p) c= Y \ X-section(E,p);
    now let y be set;
     assume A5: y in Y \ X-section(E,p); then
     y in Y & not y in X-section(E,p) by XBOOLE_0:def 5; then
A6:  not [p,y] in E;
     [p,y] in [:X,Y:] by A1,A5,ZFMISC_1:def 2; then
     [p,y] in [:X,Y:] \ E by A6,XBOOLE_0:def 5;
     hence y in X-section([:X,Y:] \ E,p) by A5;
    end; then
    Y \ X-section(E,p) c= X-section([:X,Y:] \ E,p);
    hence X-section([:X,Y:] \ E,p) = Y \ X-section(E,p) by A4;
   end;
   assume A7: p in Y;
   now let y be set;
    assume A8: y in Y-section([:X,Y:] \ E,p); then
A9:  ex y1 be Element of X st y = y1 & [y1,p] in [:X,Y:] \ E;
    now assume y in Y-section(E,p); then
     ex y2 be Element of X st y = y2 & [y2,p] in E;
     hence contradiction by A9,XBOOLE_0:def 5;
    end;
    hence y in X \ Y-section(E,p) by A8,XBOOLE_0:def 5;
   end; then
A10:Y-section([:X,Y:] \ E,p) c= X \ Y-section(E,p);
   now let y be set;
    assume A11: y in X \ Y-section(E,p); then
    y in X & not y in Y-section(E,p) by XBOOLE_0:def 5; then
A12:not [y,p] in E;
    [y,p] in [:X,Y:] by A7,A11,ZFMISC_1:def 2; then
    [y,p] in [:X,Y:] \ E by A12,XBOOLE_0:def 5;
    hence y in Y-section([:X,Y:] \ E,p) by A11;
   end; then
   X \ Y-section(E,p) c= Y-section([:X,Y:] \ E,p);
   hence Y-section([:X,Y:] \ E,p) = X \ Y-section(E,p) by A10;
end;
