
theorem Th29:
for X,Y be non empty set, E1,E2 be Subset of [:X,Y:], p be set
 st E1 misses E2
 holds X-section(E1,p) misses X-section(E2,p)
   & Y-section(E1,p) misses Y-section(E2,p)
proof
   let X,Y be non empty set, E1,E2 be Subset of [:X,Y:], p be set;
   assume A1: E1 misses E2;
   now let q be set;
    assume q in X-section(E1,p) /\ X-section(E2,p); then
A2: q in X-section(E1,p) & q in X-section(E2,p) by XBOOLE_0:def 4; then
A3: ex y1 be Element of Y st q = y1 & [p,y1] in E1;
    ex y2 be Element of Y st q = y2 & [p,y2] in E2 by A2;
    hence contradiction by A1,A3,XBOOLE_0:def 4;
   end; then
   X-section(E1,p) /\ X-section(E2,p) is empty;
   hence X-section(E1,p) misses X-section(E2,p);
   now let q be set;
    assume q in Y-section(E1,p) /\ Y-section(E2,p); then
A4: q in Y-section(E1,p) & q in Y-section(E2,p) by XBOOLE_0:def 4; then
A5: ex x1 be Element of X st q = x1 & [x1,p] in E1;
    ex x2 be Element of X st q = x2 & [x2,p] in E2 by A4;
    hence contradiction by A1,A5,XBOOLE_0:def 4;
   end; then
   Y-section(E1,p) /\ Y-section(E2,p) is empty;
   hence Y-section(E1,p) misses Y-section(E2,p);
end;
