
theorem Th24:
for X be set, Y be non empty set, p be set, F be SetSequence of [:X,Y:],
 Fy be SetSequence of Y st
  ( for n be Nat holds Fy.n = X-section(F.n,p) )
 holds X-section(union rng F,p) = union rng Fy
proof
   let X be set, Y be non empty set, p be set, F be SetSequence of [:X,Y:],
    Fy be SetSequence of Y;
   assume A2: for n be Nat holds Fy.n = X-section(F.n,p);
   now let q be set;
    assume q in X-section(union rng F,p); then
    consider y1 be Element of Y such that
A3:  q = y1 & [p,y1] in union rng F;
    consider T be set such that
A4:  [p,y1] in T & T in rng F by A3,TARSKI:def 4;
    consider m be Element of NAT such that
A5:  T = F.m by A4,FUNCT_2:113;
    Fy.m = X-section(F.m,p) by A2; then
    q in Fy.m & Fy.m in rng Fy by A3,A4,A5,FUNCT_2:112;
    hence q in union rng Fy by TARSKI:def 4;
   end; then
A7:X-section(union rng F,p) c= union rng Fy;
   now let q be set;
    assume q in union rng Fy; then
    consider T be set such that
A8:  q in T & T in rng Fy by TARSKI:def 4;
    consider m be Element of NAT such that
A9:  T = Fy.m by A8,FUNCT_2:113;
    q in X-section(F.m,p) by A2,A8,A9; then
    consider y1 be Element of Y such that
A10: q = y1 & [p,y1] in F.m;
    F.m in rng F by FUNCT_2:112; then
    [p,y1] in union rng F by A10,TARSKI:def 4;
    hence q in X-section(union rng F,p) by A10;
   end; then
   union rng Fy c= X-section(union rng F,p);
   hence X-section(union rng F,p) = union rng Fy by A7;
end;
