
theorem Th16:
for X,Y be non empty set, A be Subset of X, B be Subset of Y, p be set
 holds
   (p in A implies X-section([:A,B:],p) = B)
 & (not p in A implies X-section([:A,B:],p) = {})
 & (p in B implies Y-section([:A,B:],p) = A)
 & (not p in B implies Y-section([:A,B:],p) = {})
proof
   let X,Y be non empty set, A be Subset of X, B be Subset of Y, p be set;
   set E = [:A,B:];
   thus p in A implies X-section([:A,B:],p) = B by RELAT_1:168;
   hereby assume A5: not p in A;
    now let y be set;
     assume y in X-section([:A,B:],p); then
     ex y1 be Element of Y st y = y1 & [p,y1] in E;
     hence contradiction by A5,ZFMISC_1:87;
    end; then
    X-section([:A,B:],p) is empty;
    hence X-section([:A,B:],p) = {};
   end;
   hereby assume A4: p in B;
    now let x be set;
     assume x in Y-section([:A,B:],p); then
     ex x1 be Element of X st x = x1 & [x1,p] in E;
     hence x in A by ZFMISC_1:87;
    end; then
A5: Y-section([:A,B:],p) c= A;
    now let x be set;
     assume A6: x in A; then
     [x,p] in [:A,B:] by A4,ZFMISC_1:87;
     hence x in Y-section([:A,B:],p) by A6;
    end; then
    A c= Y-section([:A,B:],p);
    hence Y-section([:A,B:],p) = A by A5;
   end;
   assume A7: not p in B;
    now let x be set;
     assume x in Y-section([:A,B:],p); then
     ex x1 be Element of X st x = x1 & [x1,p] in E;
     hence contradiction by A7,ZFMISC_1:87;
    end; then
    Y-section([:A,B:],p) is empty;
    hence Y-section([:A,B:],p) = {};
end;
