
theorem Th26:
for X be non empty set, Y be set, p be set, F be SetSequence of [:X,Y:],
 Fx be SetSequence of X st
  ( for n be Nat holds Fx.n = Y-section(F.n,p) )
 holds Y-section(union rng F,p) = union rng Fx
proof
   let X be non empty set, Y be set, p be set, F be SetSequence of [:X,Y:],
    Fx be SetSequence of X;
   assume A2: for n be Nat holds Fx.n = Y-section(F.n,p);
   now let q be set;
    assume q in Y-section(union rng F,p); then
    consider x1 be Element of X such that
A3:  q = x1 & [x1,p] in union rng F;
    consider T be set such that
A4:  [x1,p] 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;
    Fx.m = Y-section(F.m,p) by A2; then
    q in Fx.m & Fx.m in rng Fx by A3,A4,A5,FUNCT_2:112;
    hence q in union rng Fx by TARSKI:def 4;
   end; then
A7:Y-section(union rng F,p) c= union rng Fx;
   now let q be set;
    assume q in union rng Fx; then
    consider T be set such that
A8:  q in T & T in rng Fx by TARSKI:def 4;
    consider m be Element of NAT such that
A9:  T = Fx.m by A8,FUNCT_2:113;
    q in Y-section(F.m,p) by A2,A8,A9; then
    consider x1 be Element of X such that
A10: q = x1 & [x1,p] in F.m;
    F.m in rng F by FUNCT_2:112; then
    [x1,p] in union rng F by A10,TARSKI:def 4;
    hence q in Y-section(union rng F,p) by A10;
   end; then
   union rng Fx c= Y-section(union rng F,p);
   hence Y-section(union rng F,p) = union rng Fx by A7;
end;
