theorem Th22: A 'U' B in rng P`1 & Q in compn P implies untn(A,B) in rng Q`1
  proof
    assume that
A1: A 'U' B in rng P`1 and
A2: Q in compn P;
    consider Q1 be Element of pairs such that
    Q = Q1 and
A3: Q1 in comp untn P by A2;
    consider x such that
    Q1 in x and
A4: x in {(comp R) where R is Element of pairs : R in untn P}
    by TARSKI:def 4,A3;
    consider R be PNPair such that
    x = comp R and
A5: R in untn P by A4;
    ex R1 be PNPair st R1 = R&rng (R1`1) = untn rng (P`1) & rng (R1`2) =
    untn rng (P`2) by A5;
    then A6: untn(A,B) in rng (R`1) by A1;
    Q is_completion_of R by A5,A2,Th19;
    then rng (R`1) c= rng (Q`1);
    hence untn(A,B) in rng Q`1 by A6;
  end;
