theorem Th5:
  f is one-to-one implies (a1 [= b1 iff (f.a1) [= (f.b1))
proof
  assume
A1: f is one-to-one;
  reconsider f as Function of L1,L2;
  f.a1 [= f.b1 implies a1 [= b1
  proof
    assume f.a1 [= f.b1;
    then f.(a1 "\/" b1) = f.b1 by D1;
    hence a1 "\/" b1 = b1 by A1,FUNCT_2:19;
  end;
  hence thesis by Th4;
end;
