
theorem Th27:
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(meet rng F,p) = meet 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(meet rng F,p); then
    consider y1 be Element of X such that
A3:  q = y1 & [y1,p] in meet rng F;
    for T be set st T in rng Fx holds q in T
    proof
     let T be set;
     assume T in rng Fx; then
     consider n be object such that
B1:   n in dom Fx & T = Fx.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
     [q,p] in F.n by A3,SETFAM_1:def 1; then
     q in Y-section(F.n,p) by A3;
     hence q in T by B1,A2;
    end;
    hence q in meet rng Fx by SETFAM_1:def 1;
   end; then
A7:Y-section(meet rng F,p) c= meet rng Fx;
   now let q be set;
    assume B0: q in meet rng Fx;
    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 Fx = NAT by FUNCT_2:def 1; then
     Fx.n in rng Fx by B2,FUNCT_1:3; then
     q in Fx.n by B0,SETFAM_1:def 1; then
     q in Y-section(F.n,p) by A2; then
     ex y be Element of X st q = y & [y,p] in F.n;
     hence [q,p] in T by B2;
    end; then
    [q,p] in meet rng F by SETFAM_1:def 1;
    hence q in Y-section(meet rng F,p) by B0;
   end; then
   meet rng Fx c= Y-section(meet rng F,p);
   hence Y-section(meet rng F,p) = meet rng Fx by A7;
end;
