
theorem Th3:
for X,Y be non empty set, f be Function of X,Y, S be SigmaField of X
  st f is bijective holds .:f.:S is SigmaField of Y
proof
    let X,Y be non empty set, f be Function of X,Y, S be SigmaField of X;
    set S1 = .:f.:S;
    assume
A1: f is bijective; then
    .:f is bijective by Th1; then
A2: dom (.:f) = bool X & rng (.:f) = bool Y by FUNCT_2:def 1,def 3;

A3: f is one-to-one & f is onto by A1;
    reconsider S1 as Field_Subset of Y by A1,Th2;

    for A1 being SetSequence of Y st rng A1 c= S1 holds Intersection A1 in S1
    proof
     let A1 be SetSequence of Y;
     assume
A4:  rng A1 c= S1;
     defpred P[set,object] means A1.$1 = (.:f).$2 & $2 in S;

A5:  for x being Element of NAT ex y being Element of bool X st P[x,y]
     proof
      let x be Element of NAT;
      reconsider B = A1.x as Subset of Y;
      B in rng A1 by FUNCT_2:4; then
      consider z be object such that
A6:   z in dom (.:f) & z in S & B = .:f.z by A4,FUNCT_1:def 6;
      reconsider z as Element of bool X by A6;
      take z;
      thus .:f.z = A1.x by A6;
      thus z in S by A6;
     end;

     consider T being Function of NAT,bool X such that
A7:  for x being Element of NAT holds P[x,T.x] from FUNCT_2:sch 3(A5);
     reconsider T as SetSequence of X;

     now let s be object;
      assume s in rng T; then
      ex n be Element of NAT st n in dom T & s=T.n by PARTFUN1:3;
      hence s in S by A7;
     end; then
A8: rng T c= S;

     for y be object holds y in f.:(Intersection T) iff y in Intersection A1
     proof
      let y be object;
      hereby assume y in f.:(Intersection T); then
       consider x be object such that
A9:    x in dom f & x in Intersection T & y = f.x by FUNCT_1:def 6;
       now let n be Nat;
A10:    x in T.n by PROB_1:13,A9;
        n in NAT by ORDINAL1:def 12; then
        A1.n = .:f.(T.n) & T.n in S by A7; then
        A1.n = f.:(T.n) by A1,Th1;
        hence y in A1.n by A9,A10,FUNCT_1:def 6;
       end;
       hence y in Intersection A1 by PROB_1:13;
      end;

      assume
A11:  y in Intersection A1; then
A12:  y in A1.0 by PROB_1:13;
A13:  A1.0 = .:f.(T.0) & T.0 in S by A7;
      .:f.(T.0) = f.:(T.0) by A1,Th1; then
      consider t0 be object such that
A14:  t0 in dom f & t0 in T.0 & y = f.t0 by A12,A13,FUNCT_1:def 6;

      now let n be Nat;
A15:   y in A1.n by A11,PROB_1:13;
       n in NAT by ORDINAL1:def 12; then
       A1.n = .:f.(T.n) by A7; then
       A1.n = f.:(T.n) by A1,Th1; then
       ex tn be object st tn in dom f & tn in T.n & y = f.tn
         by A15,FUNCT_1:def 6;
       hence t0 in T.n by A3,A14;
      end; then
      t0 in Intersection T by PROB_1:13;
      hence y in f.:(Intersection T) by A14,FUNCT_1:def 6;
     end; then
     f.:(Intersection T) = Intersection A1 by TARSKI:2; then
A16: .:f.(Intersection T) = Intersection A1 by A1,Th1;
     Intersection T in S by A8,PROB_1:def 6;
     hence Intersection A1 in S1 by A2,A16,FUNCT_1:def 6;
    end; then
    S1 is sigma-multiplicative;
    hence thesis;
end;
