theorem
  'not' p.(x,y) = 'not' (p.(x,y)) & (QuantNbr(p) = QuantNbr(p.(x,y))
  implies QuantNbr('not' p) = QuantNbr('not' p.(x,y)))
proof
  set S = ['not' p,Sbst(x,y)];
A1: S = Sub_not [p,Sbst(x,y)] by Th16;
  then
A2: ('not' p).(x,y) = 'not' CQC_Sub([p,Sbst(x,y)]) by SUBSTUT1:29;
  QuantNbr(p) = QuantNbr(p.(x,y)) implies QuantNbr('not' p) = QuantNbr((
  'not' p).(x,y))
  proof
    assume
A3: QuantNbr(p) = QuantNbr(p.(x,y));
    QuantNbr(('not' p).(x,y)) = QuantNbr(p.(x,y)) by A2,CQC_SIM1:16;
    hence thesis by A3,CQC_SIM1:16;
  end;
  hence thesis by A1,SUBSTUT1:29;
end;
