
theorem MMS1:
  for n be Element of NAT st 0 < n holds
    (Nat2BL.n).(LenBSeq n) = 1
  proof
    let n be Element of NAT;
    assume AS: 0 < n;
    assume AC: (Nat2BL.n).(LenBSeq n) <> 1;
    L1: Nat2BL.n = (LenBSeq n) -BinarySequence n by BINARI_6:def 2;
    reconsider x = Nat2BL.n as Element of BOOLEAN*;
    LAC: not x in {y where y is Element of BOOLEAN*: y.(len y) =1 }
    proof
      assume x in {y where y is Element of BOOLEAN*: y.(len y) =1 };
      then ex y be Element of BOOLEAN* st x=y & y.(len y) =1;
      hence contradiction by AC,L1,FINSEQ_3:153;
    end;
    0 in NAT; then
    PO: 0 in dom Nat2BL by FUNCT_2:def 1;
    n in NAT; then
    TLL: n in dom Nat2BL by FUNCT_2:def 1; then
    x in rng Nat2BL by FUNCT_1:3; then
    x in {y where y is Element of BOOLEAN*: y.(len y) =1 }
    or x in { <* 0 *> } by BINARI_6:11,XBOOLE_0:def 3; then
    Nat2BL.n = <* 0 *> by TARSKI:def 1,LAC;
    hence contradiction by AS,TLL,PO,zerobs, BINARI_6:12;
  end;
