theorem Th53:
  dom q1 misses dom q2 implies dom Shift(q1,i) misses dom Shift(q2,i)
proof
  assume
A1: dom q1 misses dom q2;
A2: dom Shift(q1,i) = {k+i where k is Nat: k in dom q1} by Def12;
A3: dom Shift(q2,i) = {k+i where k is Nat: k in dom q2} by Def12;
  now
    given x being object such that
A4: x in dom Shift(q1,i) /\ dom Shift(q2,i);
A5: x in dom Shift(q1,i) by A4,XBOOLE_0:def 4;
A6: x in dom Shift(q2,i) by A4,XBOOLE_0:def 4;
A7: ex k1 being Nat st ( x = k1+i)&( k1 in dom q1) by A2,A5;
    consider k2 being Nat such that
A8: x = k2+i and
A9: k2 in dom q2 by A3,A6;
    k2 in dom q1 /\ dom q2 by A7,A8,A9,XBOOLE_0:def 4;
    hence contradiction by A1;
  end;
  hence dom Shift(q1,i) /\ dom Shift(q2,i) = {} by XBOOLE_0:def 1;
end;
