theorem Th27:
  f is one-to-one implies (f.i "/\" f.k [= f.j implies f.k [= f.(i => j))
proof
  assume
A1: f is one-to-one;
  hereby
    assume f.i "/\" f.k [= f.j;
    then f.(i "/\" k) [= f.j by D2;
    then i "/\" k [= j by A1,Th5;
    then k [= (i => j) by FILTER_0:def 7;
    hence f.k [= f.(i => j) by Th4;
  end;
end;
