reserve i,j,k,n for Nat;
reserve x,x1,x2,x3,y1,y2,y3 for set;

theorem Th11:
  for R,S,T be non empty RelStr holds R embeds S & S embeds T
  implies R embeds T
proof
  let R,S,T be non empty RelStr;
  assume R embeds S;
  then consider f being Function of S,R such that
A1: f is one-to-one and
A2: for x,y being Element of S holds [x,y] in the InternalRel of S iff [
  f.x,f.y] in the InternalRel of R;
  assume S embeds T;
  then consider g being Function of T,S such that
A3: g is one-to-one and
A4: for x,y being Element of T holds [x,y] in the InternalRel of T iff [
  g.x,g.y] in the InternalRel of S;
  reconsider h=f*g as Function of T,R;
  take h;
  thus h is one-to-one by A1,A3;
  thus for x,y being Element of T holds [x,y] in the InternalRel of T iff [h.x
  ,h.y] in the InternalRel of R
  proof
    let x,y be Element of T;
    thus [x,y] in the InternalRel of T implies [h.x,h.y] in the InternalRel of
    R
    proof
      assume [x,y] in the InternalRel of T;
      then
A5:   [g.x,g.y] in the InternalRel of S by A4;
      h.x=f.(g.x) & h.y=f.(g.y) by FUNCT_2:15;
      hence thesis by A2,A5;
    end;
    thus [h.x,h.y] in the InternalRel of R implies [x,y] in the InternalRel of
    T
    proof
A6:   h.x=f.(g.x) & h.y=f.(g.y) by FUNCT_2:15;
      assume [h.x,h.y] in the InternalRel of R;
      then [g.x,g.y] in the InternalRel of S by A2,A6;
      hence thesis by A4;
    end;
  end;
end;
