
theorem Th29:
for X,Y be non empty set, Z be set, f be PartFunc of [:X,Y:],Z,
 x be Element of X, y be Element of Y holds
 rng(ProjPMap1(f,x)) c= rng f & rng(ProjPMap2(f,y)) c= rng f
proof
    let X,Y be non empty set, Z be set, f be PartFunc of [:X,Y:],Z,
    x be Element of X, y be Element of Y;
    now let z be Element of Z;
     assume z in rng(ProjPMap1(f,x)); then
     consider y be Element of Y such that
A1:   y in dom(ProjPMap1(f,x)) & z = (ProjPMap1(f,x)).y by PARTFUN1:3;
     y in X-section(dom f,x) by A1,MESFUN12:def 3; then
     y in {y where y is Element of Y: [x,y] in dom f} by MEASUR11:def 4;
     then
A2:  ex y2 be Element of Y st y2 = y & [x,y2] in dom f; then
     z = f.(x,y) by A1,MESFUN12:def 3;
     hence z in rng f by A2,FUNCT_1:3;
    end;
    hence rng(ProjPMap1(f,x)) c= rng f;

    now let z be Element of Z;
     assume z in rng(ProjPMap2(f,y)); then
     consider x be Element of X such that
A3:   x in dom(ProjPMap2(f,y)) & z = (ProjPMap2(f,y)).x by PARTFUN1:3;
     x in Y-section(dom f,y) by A3,MESFUN12:def 4; then
     x in {x where x is Element of X: [x,y] in dom f} by MEASUR11:def 5;
     then
A4:  ex x2 be Element of X st x2 = x & [x2,y] in dom f; then
     z = f.(x,y) by A3,MESFUN12:def 4;
     hence z in rng f by A4,FUNCT_1:3;
    end;
    hence thesis;
end;
