
theorem Th25:
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(meet rng F,p) = meet 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(meet rng F,p); then
    consider y1 be Element of Y such that
A3:  q = y1 & [p,y1] in meet rng F;
    for T be set st T in rng Fy holds q in T
    proof
     let T be set;
     assume T in rng Fy; then
     consider n be object such that
B1:   n in dom Fy & T = Fy.n by FUNCT_1:def 3;
     reconsider n as Element of NAT by B1;
     dom F = NAT by FUNCT_2:def 1; then
     F.n in rng F by FUNCT_1:3; then
     [p,q] in F.n by A3,SETFAM_1:def 1; then
     q in X-section(F.n,p) by A3;
     hence q in T by B1,A2;
    end;
    hence q in meet rng Fy by SETFAM_1:def 1;
   end; then
A7:X-section(meet rng F,p) c= meet rng Fy;
   now let q be set;
    assume B0: q in meet rng Fy;
    now let T be set;
     assume T in rng F; then
     consider n be object such that
B2:   n in dom F & T = F.n by FUNCT_1:def 3;
     reconsider n as Nat by B2;
     dom Fy = NAT by FUNCT_2:def 1; then
     Fy.n in rng Fy by B2,FUNCT_1:3; then
     q in Fy.n by B0,SETFAM_1:def 1; then
     q in X-section(F.n,p) by A2; then
     ex y be Element of Y st q = y & [p,y] in F.n;
     hence [p,q] in T by B2;
    end; then
    [p,q] in meet rng F by SETFAM_1:def 1;
    hence q in X-section(meet rng F,p) by B0;
   end; then
   meet rng Fy c= X-section(meet rng F,p);
   hence X-section(meet rng F,p) = meet rng Fy by A7;
end;
