reserve n for non zero Nat,
  j,k,l,m for Nat,
  g,h,i for Integer;

theorem Th33:
  ex x be Element of BOOLEAN st 2sComplement(m+1,i) = 2sComplement (m,i)^<*x*>
proof
  consider a be Element of m-tuples_on BOOLEAN,
   b be Element of BOOLEAN such that
A1: 2sComplement(m+1,i) = a^<*b*> by FINSEQ_2:117;
  now
    per cases;
    suppose
      m > 0;
      then reconsider m as non zero Nat;
      for j be Nat st j in Seg m holds 2sComplement(m,i).j = a.j
      proof
A2:     len 2sComplement(m,i) = m by CARD_1:def 7;
        let j be Nat such that
A3:     j in Seg m;
        reconsider j as Nat;
A4:     1 <= j by A3,FINSEQ_1:1;
        len a = m by CARD_1:def 7;
        then
A5:     j in dom a by A3,FINSEQ_1:def 3;
A6:     j <= m by A3,FINSEQ_1:1;
        m <= m + 1 by NAT_1:11;
        then len 2sComplement(m+1,i) = m+1 & j <= m + 1 by A6,CARD_1:def 7
,XXREAL_0:2;
        then 2sComplement(m+1,i).j = 2sComplement(m+1,i)/.j by A4,FINSEQ_4:15
          .= 2sComplement(m,i)/.j by A4,A6,Th32
          .= 2sComplement(m,i).j by A2,A4,A6,FINSEQ_4:15;
        hence thesis by A1,A5,FINSEQ_1:def 7;
      end;
      then a = 2sComplement(m,i) by FINSEQ_2:119;
      hence thesis by A1;
    end;
    suppose
A7:   m = 0;
      then consider c be Element of BOOLEAN such that
A8:   2sComplement(m+1,i) = <*c*> by FINSEQ_2:97;
A9:   2sComplement(m+1,i) = {}^<*c*> by A8,FINSEQ_1:34;
      2sComplement(m,i) = {} by A7;
      hence thesis by A9;
    end;
  end;
  hence thesis;
end;
