theorem
  (p '&' q).(x,y) = (p.(x,y)) '&' (q.(x,y)) & ( QuantNbr(p) = QuantNbr(p
.(x,y)) & QuantNbr(q) = QuantNbr(q.(x,y)) implies QuantNbr(p '&'q) = QuantNbr((
  p '&' q).(x,y)))
proof
  set S = [p '&' q,Sbst(x,y)];
  set S1 = [p,Sbst(x,y)];
  set S2 = [q,Sbst(x,y)];
A1: S1`2 = Sbst(x,y) & S2`2 = Sbst(x,y);
  S = CQCSub_&(S1,S2) by Th19;
  then
A2: S = Sub_&(S1,S2) by A1,SUBLEMMA:def 3;
  then
A3: (p '&' q).(x,y) = (CQC_Sub(S1)) '&' (CQC_Sub(S2)) by A1,SUBSTUT1:31;
  QuantNbr(p) = QuantNbr(p.(x,y)) & QuantNbr(q) = QuantNbr(q.(x,y))
  implies QuantNbr(p '&' q) = QuantNbr((p '&' q).(x,y))
  proof
    assume
A4: QuantNbr(p) = QuantNbr(p.(x,y)) & QuantNbr(q) = QuantNbr(q.(x,y));
    QuantNbr((p '&' q).(x,y)) = QuantNbr(p.(x,y)) + QuantNbr(q.(x,y)) by A3,
CQC_SIM1:17;
    hence thesis by A4,CQC_SIM1:17;
  end;
  hence thesis by A1,A2,SUBSTUT1:31;
end;
