
theorem Th17:
for X,Y be non empty set, E be Subset of [:X,Y:], p be set holds
  ( not p in X implies X-section(E,p) = {} )
& ( not p in Y implies Y-section(E,p) = {} )
proof
   let X,Y be non empty set, E be Subset of [:X,Y:], p be set;
   hereby assume A1: not p in X;
    now let y be set;
     assume y in X-section(E,p); then
     ex y1 be Element of Y st y = y1 & [p,y1] in E;
     hence contradiction by A1,ZFMISC_1:87;
    end; then
    X-section(E,p) is empty;
    hence X-section(E,p) = {};
   end;
   assume A7: not p in Y;
   now let y be set;
    assume y in Y-section(E,p); then
    ex y1 be Element of X st y = y1 & [y1,p] in E;
    hence contradiction by A7,ZFMISC_1:87;
   end; then
   Y-section(E,p) is empty;
   hence Y-section(E,p) = {};
end;
