
theorem Th23:
for X be non empty set, Y be set, F be FinSequence of bool [:X,Y:],
 Fx be FinSequence of bool X, p be set st
  dom F = dom Fx
& ( for n be Nat st n in dom Fx 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, F be FinSequence of bool [:X,Y:],
   Fx be FinSequence of bool X, p be set;
   assume that
A1: dom F = dom Fx and
A2: for n be Nat st n in dom Fx holds Fx.n = Y-section(F.n,p);
   now let q be set;
    assume q in Y-section(union rng F,p); then
    consider q1 be Element of X such that
A3:  q = q1 & [q1,p] in union rng F;
    consider T be set such that
A4:  [q1,p] in T & T in rng F by A3,TARSKI:def 4;
    consider m be Element of NAT such that
A5:  m in dom F & T = F.m by A4,PARTFUN1:3;
    Fx.m = Y-section(F.m,p) by A1,A2,A5; then
    q in Fx.m & Fx.m in rng Fx by A1,A3,A4,A5,FUNCT_1:3;
    hence q in union rng Fx by TARSKI:def 4;
   end; then
A6: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
A7:  q in T & T in rng Fx by TARSKI:def 4;
    consider m be Element of NAT such that
A8:  m in dom Fx & T = Fx.m by A7,PARTFUN1:3;
    q in Y-section(F.m,p) by A2,A7,A8; then
    consider q1 be Element of X such that
A9:  q = q1 & [q1,p] in F.m;
    F.m in rng F by A1,A8,FUNCT_1:3; then
    [q1,p] in union rng F by A9,TARSKI:def 4;
    hence q in Y-section(union rng F,p) by A9;
   end; then
   union rng Fx c= Y-section(union rng F,p);
   hence Y-section(union rng F,p) = union rng Fx by A6;
end;
