
theorem Th22:
for X be set, Y be non empty set, F be FinSequence of bool [:X,Y:],
 Fy be FinSequence of bool Y, p be set st
  dom F = dom Fy
& ( for n be Nat st n in dom Fy 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, F be FinSequence of bool [:X,Y:],
   Fy be FinSequence of bool Y, p be set;
   assume that
A1: dom F = dom Fy and
A2: for n be Nat st n in dom Fy holds Fy.n = X-section(F.n,p);
   now let q be set;
    assume q in X-section(union rng F,p); then
    consider q1 be Element of Y such that
A3:  q = q1 & [p,q1] in union rng F;
    consider T be set such that
A4:  [p,q1] 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;
    Fy.m = X-section(F.m,p) by A1,A2,A5; then
    q in Fy.m & Fy.m in rng Fy by A1,A3,A4,A5,FUNCT_1:3;
    hence q in union rng Fy by TARSKI:def 4;
   end; then
A6: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
A7:  q in T & T in rng Fy by TARSKI:def 4;
    consider m be Element of NAT such that
A8:  m in dom Fy & T = Fy.m by A7,PARTFUN1:3;
    q in X-section(F.m,p) by A2,A7,A8; then
    consider q1 be Element of Y such that
A9:  q = q1 & [p,q1] in F.m;
    F.m in rng F by A1,A8,FUNCT_1:3; then
    [p,q1] in union rng F by A9,TARSKI:def 4;
    hence q in X-section(union rng F,p) by A9;
   end; then
   union rng Fy c= X-section(union rng F,p);
   hence X-section(union rng F,p) = union rng Fy by A6;
end;
